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MATHS SYMBOLS - EXPONENTIALS + LOGARITHMS and THEIR PROPERTIES

Ist. Superiore Marini-Gioia - Enzo Exposyto, profile picture
Ist. Superiore Marini-Gioia - Enzo Exposyto

The document provides a comprehensive overview of exponentials and logarithms, including definitions, properties, and proofs. It explains how logarithms and exponentials relate as inverse operations and includes numerous examples of logarithmic calculations and their properties. Furthermore, it details specific operations such as the logarithm of a power, product, and root, along with the change of base and proof of properties.

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ENZO EXPOSYTO MATHS SYMBOLS PROPERTIES of EXPONENTIALS and LOGARITHMS
 Enzo Exposyto 1
2X ex 2-x e-x EXPONENTIALS LOGARITHMS log2(x) ln(x) log(x)
 Enzo Exposyto 2

 Enzo Exposyto 3
1 - Exponential - Definition 6 2 - Exponentials - Their Properties 8 3 - Exponentials and Logarithms 15 4 - Logarithm - Definition and Examples 25 5 - Logarithms - Their Properties 34 6 - log(y) and ln(y) - Properties 46 Enzo Exposyto 4
7 - logb(bx) = x - Proofs 61 8 - ylogb(y) = y - Proofs 64 9 - log of a Power - Proofs 68 10 - log of a Root - Proofs 71 11 - log of a Product - Proofs 74 12 - log of a Quotient - Proofs 77 13 - Change of Base - Proofs 82 14 - zlogb(y) = ylogb(z) - Proof 91 15 - SitoGraphy 93 Enzo Exposyto 5
EXPONENTIAL - DEFINITION Enzo Exposyto 6
EXPONENTIAL - DEFINITION DEFINITION Examples DEFINITION bx = y ... 2-1 = 1 2 ... 20 = 1 ... 21 = 2 ... b > 0 and b < > 1 (*) x element of R y > 0 "b" is the fixed base x is the variable exponent (*) The symbol < > and ≠ have the same meaning Enzo Exposyto 7
EXPONENTIALS - THEIR PROPERTIES
 Enzo Exposyto 8
EXPONENTIALS - THEIR PROPERTIES - 1 [b, c > 0 and b, c < > 1] [x, y elements of R] Z+ = {1, 2, 3, …} Property Exponentials Exponents Exponentials Exponents Result 1st bx · by = bx + y 23 · 22 = 23 + 2 = 32 2nd bx = by bx -y 23 = 22 23 -2 = 2 3rd (bx)y = bx *y (23)2 = 23 *2 = 64 4th n Z+ n√bx = bx : n 2√24 = 24 : 2 = 4 5th bx · cx = (b · c)x 22 · 32 = (2 · 3)2 = 36 6th bx ———- = cx (_b_)x c 43 ———- = 23 (_4_)3 = 2 8 Enzo Exposyto 9
EXPONENTIALS - THEIR PROPERTIES - 2 [b, c > 0 and b, c < > 1] [x, y elements of R] Z+ = {1, 2, 3, …} Property Exponents Exponentials Exponents Exponentials Result 1st bx + y = bx · by 23 + 2 = 23 · 22 = 32 2nd bx -y = bx ———- by 23 -2 = 23 = 22 2 3rd bx *y = (bx)y 23 *2 = (23)2 = 64 4th n Z+ bx : n = n√bx 24 : 2 = 2√24 = 4 5th (b · c)x = bx · cx (2 · 3)2 = 22 · 32 = 36 6th (_b_)x = c bx ———- cx (_4_)3 = 2 43 ———- = 23 8 Enzo Exposyto 10
Exponentials - 6 Properties - Proofs/Examples PROOFS / EXAMPLES [b, c > 0 and b, c < > 1] 1st b3 · b2 = (b · b · b) · (b · b) = b · b · b · b · b = b5 = b3 + 2 2nd b3 = (b · b · b) = b = b1 = b3 - 2 b2 (b · b) 3rd (b3)2 = (b · b · b) · (b · b · b) = b · b · b · b · b · b = b6 = b3 * 2 4th 2√b4 = 2√(b · b · b · b) = b · b = b2 = b4 : 2 5th b3 · c3 = (b · b · b) · (c · c · c) = b · c · b · c · b · c = … = (b·c)3 6th b3 (b · b · b) b b b b ——- = ————— = —— . —— . —— = … = (—)3 c3 (c · c · c) c c c c Enzo Exposyto 11
Exponentials - 2nd property - b superscript 0 EXPONENTIALS - THEIR PROPERTIES - 3 Reference Property Notice Proof Example 2nd Property b0 = 1 b > 0 b ≠ 1 x R b0 = bx - x 20 = 23 - 3 = bx bx = 23 23 = 1 = 1 1 = b0 b > 0 b ≠ 1 1 = 20 Enzo Exposyto 12
Exponentials - 2nd property - b superscript (-x) EXPONENTIALS - THEIR PROPERTIES - 4 Reference Property Notice Proof Example 2nd Property b-x = 1 bx b > 0 b ≠ 1 x R b-x = b0-x 2-3 = 20-3 = b0 bx = 20 23 = 1 bx = 1 23 1 = b-x bx b > 0 b ≠ 1 x R 1 = 2-3 23 Enzo Exposyto 13
Exponentials - 2nd property - Reciprocal of bx EXPONENTIALS and THEIR PROPERTIES - 5 Reference Property Notice Proof Example 2nd Property bx = 1 b-x b > 0 b ≠ 1 x R 23 = 1 2-3 1 = bx b-x b > 0 b ≠ 1 x R 1 = 23 2-3 Enzo Exposyto 14
EXPONENTIALS and LOGARITHMS
 Enzo Exposyto 15
EXPONENTIAL BASE 2: 23 = 2 × 2 x 2 = 8 The LOGARITHM BASE 2 OF 8 goes the OTHER WAY: Enzo Exposyto 16
The logarithm base 2 of 8 is 3, BECAUSE 2 cubed is 8; so the logarithm base 2 of 8 is 3: Enzo Exposyto 17
THE LOGARITHM BASE 2 OF 8 IS THE OPERATION THAT ALLOWS US OF GOING BACK TO THE EXPONENT 3 EXPONENTIAL 2x and LOGARITHM BASE 2 EXPONENT x 2x 1 2 2 4 3 8 4 16 Enzo Exposyto 18
In other words … THE LOGARITHM BASE 2 OF 8 IS THE EXPONENT 3 … MORE PRECISELY … THE LOGARITHM "3" IS THE EXPONENT WHICH WE HAVE TO PUT ON THE BASE “2” TO GET “8” Enzo Exposyto 19
Now, since log2(8) = 3 and 8 = 23 then log2(23) = 3 We can see that THE LOGARITHM "3" IS THE EXPONENT WHICH WE HAVE TO PUT ON THE BASE “2” TO GET “23”
 Enzo Exposyto 20
EXPONENTIAL and LOGARITHM with the same base CANCEL EACH OTHER. This is true because exponential and logarithm with the same base are INVERSE OPERATIONS It is just like Addition and Subtraction, Multiplication and Division, Exponentiation and Root, … when they're INVERSE OPERATIONS Enzo Exposyto 21
Now, we can introduce the ANTILOGARITHM BASE b: antilogb(x) = bx It's, simply, an EXPONENTIAL and represents the antilogarithm when we operate with a logarithm … It’s such that logb(antilogb(x)) = x The meaning is logb(bx) = x Enzo Exposyto 22
And, of course, antilogb(logb(y)) = y The meaning is blogb(y) = y Enzo Exposyto 23
These phrases - with 8 - are equivalent log2(8) = 3 < = > 23 = 8 OR 23 = 8 < = > log2(8) = 3 These phrases - with 23 - are equivalent log2(23) = 3 < = > 23 = 23 OR 23 = 23 < = > log2(23) = 3 Enzo Exposyto 24
LOGARITHM - DEFINITION and EXAMPLES Enzo Exposyto 25
LOGARITHM - DEFINITION DEFINITION Example DEFINITION logb(y) = x iff (if and only if) bx = y log2(8) = 3 because 23 = 8 b > 0 and b < > 1 y > 0 x element of R 2 > 0 and 2 < > 1 8 > 0 3 element of R The logarithm "x" is the exponent which we have to put on the base “b” to get “y” The logarithm "3" is the exponent which we have to put on the base “2” to get “8” Enzo Exposyto 26
iff bx = y
Enzo Exposyto 27
LOGARITHMS - EXAMPLES - 1 log DEFINITION because log3(9) = 2 The logarithm "2" is the exponent which we have to put on the base “3” to get “9” log3(9) = 2 because 32 = 9 log3(27) = 3 The logarithm "3" is the exponent which we have to put on the base “3” to get “27” log3(27) = 3 because 33 = 27 log4(4) = 1 The logarithm "1" is the exponent which we have to put on the base “4” to get “(4)” log4(4) = 1 because 41 = (4) Enzo Exposyto 28
LOGARITHMS - EXAMPLES - 2 log DEFINITION because log4(16) = 2 The logarithm "2" is the exponent which we have to put on the base “4” to get “16” log4(16) = 2 because 42 = 16 log5(25) = 2 The logarithm "2" is the exponent which we have to put on the base “5” to get “25” log5(25) = 2 because 52 = 25 log10(100) = 2 The logarithm "2" is the exponent which we have to put on the base “10” to get “100” log10(100) = 2 because 102 = 100 Enzo Exposyto 29
LOGARITHMS - EXAMPLES - 3 log DEFINITION because log4(1) = -1 4 The logarithm "-1" is the exponent which we have to put on the base “4” to get “1” 4 log4(1) = -1 4 because 4-1 = 1 4 log10( 1 ) = -1 10 The logarithm "-1" is the exponent which we have to put on the base “10” to get “ 1 ” 10 log10( 1 ) = -1 10 because 10-1 = 1 10 Enzo Exposyto 30
LOGARITHMS - EXAMPLES - 4 log DEFINITION because log1/2(2) = -1 The logarithm "-1" is the exponent which we have to put on the base “1” 2 to get “2” log1/2(2) = -1 because (1)-1 = 2 2 log1/4(4) = -1 The logarithm "-1" is the exponent which we have to put on the base “1” 4 to get “4” log1/4(4) = -1 because (1)-1 = 4 4 Enzo Exposyto 31
LOGARITHMS - EXAMPLES - 5 log DEFINITION because log1/2(4) = -2 The logarithm "-2" is the exponent which we have to put on the base “1” 2 to get “4” log1/2(4) = -2 because (1)-2 = 4 2 log1/3(9) = -2 The logarithm "-2" is the exponent which we have to put on the base “1” 3 to get “9” log1/3(9) = -2 because (1)-2 = 9 3 Enzo Exposyto 32
LOGARITHMS - EXAMPLES - 6 e = 2.718281828… log DEFINITION because loge(e) = 1 The logarithm "1" is the exponent which we have to put on the base “e” to get “(e)” loge(e) = 1 because e1 = (e) loge(1) = 0 The logarithm "0" is the exponent which we have to put on the base “e” to get “1” loge(1) = 0 because e0 = 1 loge(e2) = 2 The logarithm "2" is the exponent which we have to put on the base “e” to get “(e2)” loge(e2) = 2 because e2 = (e2) Enzo Exposyto 33
LOGARITHMS - THEIR PROPERTIES
 Enzo Exposyto 34
LOGARITHMS - THEIR PROPERTIES - 1 Name Property Example base = 0 log0(y) is undefined log0(2) is undefined y > 0 For any x R-{0}, 0x ≠ 2 (really 0x = 0) and, then, log0(2) Does Not Exist base = 1 log1(y) is undefined log1(2) is undefined y > 0 For any x R, 1x ≠ 2 (really 1x = 1) and, then, log1(2) Does Not Exist logb(0) logb(0) is undefined log2(0) is undefined b > 0 and b < > 1 For any x R, 2x ≠ 0 (really 2x > 0) and, then, log2(0) Does Not Exist Enzo Exposyto 35
LOGARITHMS - THEIR PROPERTIES - 2 Name Property Examples logb(1) logb(1) = 0 log2(1) = 0 because 20 = 1b > 0 and b < > 1 logb(b) logb(b) = 1 log4(4) = 1 because 41 = (4)b > 0 and b < > 1 Enzo Exposyto 36
LOGARITHMS - THEIR PROPERTIES - 3 Name Property Example logb(bx) logb(bx) = x log2(23) = 3 because 23 = (23) b > 0 and b < > 1 x element of R blogb(y) blogb(y) = y 2log2(8) = 8 because 23 = 8 b > 0 and b < > 1 y > 0 Enzo Exposyto 37
LOGARITHMS - THEIR PROPERTIES - 4 Name Property Examples log of a Power logb(yz) = z logb(y) log2(42) = 2 log2(4) b > 0 and b < > 1 y > 0 z element of R b = 2 y = 4 z = 2 log of a Reciprocal logb(1) = logb(y-1) = - logb(y) y log2(1) = log2(4-1) = - log2(4) 4 b > 0 and b < > 1 y > 0 b = 2 y = 4 Enzo Exposyto 38
LOGARITHMS - THEIR PROPERTIES - 5 Name Property Examples log of a Root - 1 logb(n√y) = logb(y) n log2(3√8) = log2(8) 3 b > 0 and b < > 1 y > 0 n Z+ Z+ = {1, 2, 3, …} b = 2 y = 8 n = 3 log of a Root - 2 logb(n√yz) = z logb(y) n log2(3√26) = 6 log2(2) 3 b > 0 and b < > 1 y > 0 z element of R n Z+ Z+ = {1, 2, 3, …} b = 2 y = 2 z = 6 n = 3 Enzo Exposyto 39
LOGARITHMS - THEIR PROPERTIES - 6 Name Property Examples log of a Product - 1 logb(y z) = logb(y) + logb(z) log2(4 2) = log2(4) + log2(2) b > 0 and b < > 1 y, z > 0 b = 2 y = 4; z = 2 log of a Product - 2 logb(yn . zp) = n.logb(y)+p.logb(z) log2(43 . 24) = 3.log2(4)+4.log2(2) b > 0 and b < > 1 y, z > 0 n, p elements of R b = 2 y = 4; z = 2 n = 3; p = 4 Enzo Exposyto 40
LOGARITHMS - THEIR PROPERTIES - 7 Name Property Examples log of a Quotient - 1 logb(y) = logb(y.z-1) = logb(y)-logb(z) z log2(8) = log2(8.4-1) = log2(8) - log2(4) 4 b > 0 and b < > 1 y, z > 0 b = 2 y = 8; z = 4 log of a Quotient - 2 logb(y) = logb(y) - logb(z) z log2(4) = log2(4) - log2(2) 2 b > 0 and b < > 1 y, z > 0 b = 2 y = 4; z = 2 log of a Quotient - 3 logb(yn) = n.logb(y)-p.logb(z) zp log2(43) = 3.log2(4)-4.log2(2) 24 b > 0 and b < > 1 y, z > 0 n, p elements of R b = 2 y = 4; z = 2 n = 3; p = 4 Enzo Exposyto 41
LOGARITHMS - THEIR PROPERTIES - 8 Name Property Examples Base Change - 1 logb(y) = logc(y) logc(b) log2(16) = log4(16) = 2 = 4 log4(2) 1 2b, c > 0 and b, c < > 1 y > 0 Base Switch - 1 logb(c) = logc(c) = 1 logc(b) logc(b) log2(4) = log4(4) = 1 log4(2) log4(2) logc(c) = 1 log4(4) = 1 b, c > 0 and b, c < > 1 b = 2; c= 4 Base Switch - 2 logb(c) * logc(b) = 1 log2(4) * log4(2) = 1 Enzo Exposyto 42
LOGARITHMS - THEIR PROPERTIES - 9 Name Property Examples Base Change - 2 logbn(y) = logb(y) n log2-3(8) = log2(8) -3 b > 0 and b < > 1 n element of R, n < > 0 y > 0 b = 2 n = -3 y = 8 Base Change - 3a n . logbn(y) = logb(y) -3 . log2-3(8) = log2(8) n element of R, n < > 0 b > 0 and b < > 1 y > 0 n = -3 b = 2 y = 8 Base Change - 3b logb(y) = n . logbn(y) log2(4) = 2 . log22(4) b > 0 and b < > 1 y > 0 n element of R, n < > 0 b = 2 y = 4 n = 2 Enzo Exposyto 43
LOGARITHMS - THEIR PROPERTIES - 10 Name Property Examples Base Change - 4 log1/b(y) = - logb(y) log1/8(8) = - log8(8) b > 0 and b < > 1 y > 0 b = 8 y = 8 Base Change - 5 logb(1) = log1/b(y) y log2(1) = log1/2(4) 4 b > 0 and b < > 1 y > 0 b = 2 y = 4 Enzo Exposyto 44
LOGARITHMS - THEIR PROPERTIES - 11 Name Property Examples zlogb(y) zlogb(y) = ylogb(z) 2log2(8) = 8log2(2) z > 0 b > 0 and b < > 1 y > 0 z = 2 b = 2 y = 8 Enzo Exposyto 45
log(y) AND ln(y) - THEIR PROPERTIES
 Enzo Exposyto 46
log(y) AND ln(y) - THEIR PROPERTIES REMARKS: • log(y) always refers to log base 10, i. e., log(y) = log10(y) Therefore, log(y) = x if and only if 10x = y Enzo Exposyto 47
• ln(y) is called the natural logarithm and is used to represent loge(y), where the irrational number e 2.718281828: ln(y) = loge(y) Therefore, ln(y) = x if and only if ex = y Enzo Exposyto 48
• Most calculators can directly compute logs base 10 and/or the natural log. For any other base it is necessary to use the change of the base formula: logb(y) = log10(y) = log(y) log2(8) = log(8) log10(b) log(b). log(2) or logb(y) = ln(y) log2(8) = ln(8) ln(b) ln(2)
 Enzo Exposyto 49
log(y) AND ln(y) - THEIR PROPERTIES - 1 Property log ln base = 0 base = 10 base = e base = 1 base = 10 base = e logb(0) log(0) is undefined ln(0) is undefined For any x R, 10x ≠ 0 (really 10x > 0) and, then, log(0) Does Not Exist For any x R, ex ≠ 0 (really ex > 0) and, then, ln(0) Does Not Exist Enzo Exposyto 50
log(x) AND ln(x) - THEIR PROPERTIES - 2 Property log ln logb(1) log(1) = 0 ln(1) = 0 because 100 = 1 because e0 = 1 logb(b) log(10) = 1 ln(e) = 1 because 101 = 10 because e1 = e Enzo Exposyto 51
log(y) AND ln(y) - THEIR PROPERTIES - 3 Property log ln logb(bx) log(10x) = x ln(ex) = x x element of R x R blogb(y) 10log(y) = y eln(y) = y y > 0 y > 0 Enzo Exposyto 52
log(y) AND ln(y) - THEIR PROPERTIES - 4 Property log ln log of a Power log(yz) = z log(y) ln(yz) = z ln(y) y > 0 z element of R y > 0 z element of R log of a Reciprocal log(1) = log(y-1) = - log(y) y ln(1) = ln(y-1) = - ln(y) y y > 0 y > 0 Enzo Exposyto 53
log(y) AND ln(y) - THEIR PROPERTIES - 5 Property log ln log of a Root - 1 log(n√y) = log(y) n ln(n√y) = ln(y) n n Z+ Z+ = {1, 2, 3, …} y > 0 n Z+ Z+ = {1, 2, 3, …} y > 0 log of a Root - 2 log(n√yz) = z log(y) n ln(n√yz) = z ln(y) n n element of Z+ Z+ = {1, 2, 3, …} y > 0 z element of R n element of Z+ Z+ = {1, 2, 3, …} y > 0 z element of R Enzo Exposyto 54
log(y) AND ln(y) - THEIR PROPERTIES - 6 Property log ln log of a Product - 1 log(y z) = log(y) + log(z) ln(y z) = ln(y) + ln(z) y, z > 0 y, z > 0 log of a Product - 2 log(yn . zp) = n.log(y)+p.log(z) ln(yn . zp) = n.ln(y)+p.ln(z) y, z > 0 n, p elements of R y, z > 0 n, p elements of R Enzo Exposyto 55
log(y) AND ln(y) - THEIR PROPERTIES - 7 Property log ln log of a Quotient - 1 log(y) = log(y.z-1) = log(y)-log(z) z ln(y) = ln(y.z-1) = ln(y)-ln(z) z y, z > 0 y, z > 0 log of a Quotient - 2 log(y) = log(y) - log(z) z ln(y) = ln(y) - ln(z) z y, z > 0 y, z > 0 log of a Quotient - 3 log(yn) = n.log(y)-p.log(z) zp ln(yn) = n.ln(y)-p.ln(z) zp y, z > 0 n, p elements of R y, z > 0 n, p elements of R Enzo Exposyto 56
log(y) AND ln(y) - THEIR PROPERTIES - 8 Property log ln Base Change - 1 log(y) = logc(y) logc(10) ln(y) = logc(y) logc(e) y > 0 c > 0 and c < > 1 y > 0 c > 0 and c < > 1 Base Switch - 1 log(c) = logc(c) = 1 logc(10) logc(10) ln(c) = logc(c) = 1 logc(e) logc(e) logc(c) = 1 logc(c) = 1 c > 0 and c < > 1 c > 0 and c < > 1 Base Switch - 2 log(c) * logc(10) = 1 ln(c) * logc(e) = 1 c > 0 and c < > 1 c > 0 and c < > 1 Enzo Exposyto 57
log10(y) AND loge(y) - THEIR PROPERTIES - 9 Property log10 loge Base Change - 2 log10n(y) = log10(y) n logen(y) = loge(y) n n element of R, n < > 0 y > 0 n element of R, n < > 0 y > 0 Base Change - 3a n . log10n(y) = log10(y) n . logen(y) = loge(y) n element of R, n < > 0 y > 0 n element of R, n < > 0 y > 0 Base Change - 3b log10(y) = n . log10n(y) loge(y) = n . logen(y) y > 0 n element of R, n < > 0 y > 0 n element of R, n < > 0 Enzo Exposyto 58
log10(y) AND loge(y) - THEIR PROPERTIES - 10 Property log10 loge Base Change - 4 log1/10(y) = - log10(y) log1/e(y) = - loge(y) y > 0 y > 0 Base Change - 5 log10(1) = log1/10(y) y loge(1) = log1/e(y) y y > 0 y > 0 Enzo Exposyto 59
log10(y) AND loge(y) - THEIR PROPERTIES - 11 Property log10 loge zlogb(y) zlog10(y) = ylog10(z) zloge(y) = yloge(z) z > 0 y > 0 z > 0 y > 0 Enzo Exposyto 60
logb(bx) = x PROOFS
 Enzo Exposyto 61
logb(bx) = x a) The log of bx is the exponent which we have to put on the base b to get bx itself and, therefore, it’s “x” b) If, from x, we ‘go’ to bx and, then, from bx, we ‘go’ to logb(bx), since logb(bx) is the inverse operation of bx, we go back to x … therefore, logb(bx) = x In other words (remembering that bx = antilogb(x)): logb(antilogb(x)) = x 
 Enzo Exposyto 62
logb(bx) = x c) Let's set bx = y and, then, we ‘do’ the logarithms in base b of both sides; we get logb(bx) = logb(y) Since bx = y <=> logb(y) = x we can write logb(bx) = logb(y) = x Therefore, logb(bx) = x Q.E.D. Enzo Exposyto 63
blogb(y) = y PROOFS 
 Enzo Exposyto 64
blogb(y) = y a) If from y, we ‘go’ to logb(y) and, then, from logb(y), we ‘go’ to blogb(y), since blogb(y) is the inverse operation of logb(y), we go back to y … therefore, blogb(y) = y In other words (remembering that blogb(y) = antilogb(logb(y))): antilogb(logb(y)) = y 
 Enzo Exposyto 65
blogb(y) = y b) Let's set bx = y and, then, we ‘do’ the log in base b of both sides; we get logb(bx) = logb(y) Remembering that (pages 62-63) logb(bx) = x we can write logb(bx) = logb(y) x = logb(y) or logb(y) = x Now, we ‘do’ the exponentials in base b of both sides and we get blogb(y) = bx Since bx = y we get blogb(y) = y Q.E.D. Enzo Exposyto 66
blogb(y) = y c) Let's set logb(y) = x and, then, on the left hand side of the equation, we get blogb(y) = bx Since logb(y) = x <=> bx = y it's blogb(y) = bx = y and, therefore, blogb(y) = y Q.E.D. Enzo Exposyto 67
log of a POWER PROOFS
 Enzo Exposyto 68
logb(yz) = z . logb(y) 1) Let's set logb(y) = l and, then, the right hand side of the equation becomes z . logb(y) = z . l 2) Since logb(y) = l <=> bl = y or y = bl the left hand side of the equation becomes logb(yz) = logb((bl)z) = logb(blz) Remembering that (pages 62-63) logb(bx) = x it's logb(blz) = lz = z . l and we can write logb(yz) = logb(blz) = z . l 3) Since the left hand side and the right hand side are equal to z . l, they are equal: logb(yz) = z . logb(y) Q.E.D. Enzo Exposyto 69
logb(1) = - logb(y) y OR - logb(y) = logb(1) y Remembering that logb(yz) = z . logb(y) [log of a Power, previous page] we get logb(1) = logb(y-1) y = (-1) . logb(y) = - logb(y) Q.E.D.
 Enzo Exposyto 70
log of a ROOT PROOFS
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logb(n√y) = logb(y) n Remembering that logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(n√y) = logb(y1/n) = 1 logb(y) n = logb(y) n Q.E.D.
 Enzo Exposyto 72
logb(n√yz) = z . logb(y) n Remembering that logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(n√yz) = logb(yz/n) = z logb(y) n = z . logb(y) n Q.E.D.
 Enzo Exposyto 73
log of a PRODUCT PROOFS
 Enzo Exposyto 74
logb(y . z) = logb(y) + logb(z) 1) Let's set logb(y) = l logb(z) = m and, then, the right hand side of the equation becomes logb(y) + logb(z) = l + m 2) Since logb(y) = l <=> bl = y or y = bl logb(z) = m <=> bm = z or z = bm the left hand side of the equation becomes logb(y . z) = logb(bl . bm) = logb(bl+m) Remembering that (pages 62-63) logb(bx) = x it's logb(bl+m) = l + m and we can write logb(y . z) = logb(bl+m) = l + m 3) Since the left hand side and the right hand side are equal to l + m, they are equal: logb(y . z) = logb(y) + logb(z) Q.E.D. Enzo Exposyto 75
logb(yn . zp) = n . logb(y) + p . logb(z) Remembering that logb(y . z) = logb(y) + logb(z) [previous page] and logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(yn . zp) = logb(yn) + logb(zp) = n . logb(y) + p . logb(z) Q.E.D.
 Enzo Exposyto 76
log of a QUOTIENT PROOFS
 Enzo Exposyto 77
logb(y) = logb(y) - logb(z) z Remembering that logb(y . z) = logb(y) + logb(z) [page 75] and logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(y) = logb(y . 1) = logb(y . z-1) z z = logb(y) + logb(z-1) = logb(y) + (-1) . logb(z) = logb(y) - logb(z) Q.E.D. Enzo Exposyto 78
logb(y) = logb(y) - logb(z) z 1) Let's set logb(y) = l logb(z) = m and, then, the right hand side of the equation becomes logb(y) - logb(z) = l - m 2) Since logb(y) = l <=> bl = y or y = bl logb(z) = m <=> bm = z or z = bm the left hand side of the equation becomes logb(y) = logb(bl ) = logb(bl-m) z bm Remembering that (pages 62-63) logb(bx) = x it's logb(bl-m) = l - m and we can write logb(y) = logb(bl-m) = l - m z Enzo Exposyto 79
logb(y) = logb(y) - logb(z) z 3) Since the left hand side and the right hand side are equal to l - m, they are equal: logb(y) = logb(y) - logb(z) z Q.E.D. Enzo Exposyto 80
logb(yn) = n . logb(y) - p . logb(z) zp Remembering that logb(y) = logb(y) - logb(z) [previous page] z and logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(yn) = logb(yn) - logb(zp) zp = n . logb(y) - p . logb(z) Q.E.D.
 Enzo Exposyto 81
CHANGE of BASE PROOFS
 Enzo Exposyto 82
logb(y) = logc(y) logc(b) 1) On the left hand side of the equation, let's set logb(y) = x 2) On the right hand side of the equation, since logb(y) = x <=> bx = y or y = bx we substitute y by bx logc(y) = logc(bx) = x . logc(b) [log of a Power, page 69] and we get logc(y) = x . logc(b) = x logc(b) logc(b) 3) Since the left hand side and the right hand side are equal to x, they are equal: logb(y) = logc(y) logc(b) Q.E.D. Enzo Exposyto 83
logb(c) = 1 logc(b) It's logb(c) = logc(c) logc(b) Since logc(c) = 1 we get logb(c) = logc(c) logc(b) = 1 logc(b) and, then, logb(c) = 1 logc(b) Q.E.D. Enzo Exposyto 84
logb(c) . logc(b) = 1 From logb(c) = 1 logc(b) multiplying both sides by logc(b), it’s logb(c) logc(b) = 1 logc(b) logc(b) and, then, logb(c) logc(b) = 1 Q.E.D. Enzo Exposyto 85
logbn(y) = logb(y) n Let's change the base bn by b: logbn(y) = logb(y) logb(bn) [log of a Power, page 69] = logb(y) n logb(b) [remember: logb(b) = 1] = logb(y) n Therefore: logbn(y) = logb(y) n Q.E.D. Enzo Exposyto 86
n . logbn(y) = logb(y) OR logb(y) = n . logbn(y) From logbn(y) = logb(y) [previous page] n multiplying both sides by n, it’s n . logbn(y) = logb(y) . n n and, then, n . logbn(y) = logb(y) Q.E.D.
 Enzo Exposyto 87
log1/b(y) = - logb(y) a) From logb(y) = n . logbn(y) [previous page] if n = -1 we get logb(y) = (-1) . logb-1(y) [remember: b-1 = 1] b logb(y) = - log1/b(y) [multiplying both sides by (-1)] - logb(y) = log1/b(y) and, then, log1/b(y) = - logb(y) Q.E.D. Enzo Exposyto 88
log1/b(y) = - logb(y) b) Let's change the base 1 by b: b log1/b(y) = logb(y) logb(1) [remember: 1 = b-1] b b = logb(y) logb(b-1) [log of a Power, page 69] = logb(y) (-1) logb(b) [remember: logb(b) = 1] = logb(y) (-1) = - logb(y) Q.E.D. Enzo Exposyto 89
logb(1) = log1/b(y) y OR log1/b(y) = logb(1) y 1) From page 70: logb(1) = - logb(y) y 2) From previous page: log1/b(y) = - logb(y) 3) Since the first and the second left hand side are equal to - logb(y), they are equal: logb(1) = log1/b(y) y Q.E.D. Enzo Exposyto 90
zlogb(y) = ylogb(z) PROOF
 Enzo Exposyto 91
zlogb(y) = ylogb(z) 1) On the left hand side of the equation, let's set logb(y) = x and we get zlogb(y) = zx 2) On the right hand side of the equation, since logb(y) = x <=> bx = y or y = bx we substitute y by bx and we get ylogb(z) = (bx)logb(z) = bxlogb(z) = blogb(zx) [log of a Power, page 69] = zx [blogb(zx) = zx See blogb(y) = y (pages 65-67)] 3) Since the left hand side and the right hand side are equal to zx, they are equal: zlogb(y) = ylogb(z) Q.E.D. 
 Enzo Exposyto 92
SitoGraphy
 Enzo Exposyto 93
https://en.m.wikipedia.org/wiki/List_of_logarithmic_identities https://www.andrews.edu/~calkins/math/webtexts/numb17.htm http://dl.uncw.edu/digilib/mathematics/algebra/mat111hb/eandl/logprop/logprop.html#sec2 Enzo Exposyto 94

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MATHS SYMBOLS - EXPONENTIALS + LOGARITHMS and THEIR PROPERTIES

  • 1. ENZO EXPOSYTO MATHS SYMBOLS PROPERTIES of EXPONENTIALS and LOGARITHMS
 Enzo Exposyto 1
  • 2. 2X ex 2-x e-x EXPONENTIALS LOGARITHMS log2(x) ln(x) log(x)
 Enzo Exposyto 2
  • 4. 1 - Exponential - Definition 6 2 - Exponentials - Their Properties 8 3 - Exponentials and Logarithms 15 4 - Logarithm - Definition and Examples 25 5 - Logarithms - Their Properties 34 6 - log(y) and ln(y) - Properties 46 Enzo Exposyto 4
  • 5. 7 - logb(bx) = x - Proofs 61 8 - ylogb(y) = y - Proofs 64 9 - log of a Power - Proofs 68 10 - log of a Root - Proofs 71 11 - log of a Product - Proofs 74 12 - log of a Quotient - Proofs 77 13 - Change of Base - Proofs 82 14 - zlogb(y) = ylogb(z) - Proof 91 15 - SitoGraphy 93 Enzo Exposyto 5
  • 7. EXPONENTIAL - DEFINITION DEFINITION Examples DEFINITION bx = y ... 2-1 = 1 2 ... 20 = 1 ... 21 = 2 ... b > 0 and b < > 1 (*) x element of R y > 0 "b" is the fixed base x is the variable exponent (*) The symbol < > and ≠ have the same meaning Enzo Exposyto 7
  • 9. EXPONENTIALS - THEIR PROPERTIES - 1 [b, c > 0 and b, c < > 1] [x, y elements of R] Z+ = {1, 2, 3, …} Property Exponentials Exponents Exponentials Exponents Result 1st bx · by = bx + y 23 · 22 = 23 + 2 = 32 2nd bx = by bx -y 23 = 22 23 -2 = 2 3rd (bx)y = bx *y (23)2 = 23 *2 = 64 4th n Z+ n√bx = bx : n 2√24 = 24 : 2 = 4 5th bx · cx = (b · c)x 22 · 32 = (2 · 3)2 = 36 6th bx ———- = cx (_b_)x c 43 ———- = 23 (_4_)3 = 2 8 Enzo Exposyto 9
  • 10. EXPONENTIALS - THEIR PROPERTIES - 2 [b, c > 0 and b, c < > 1] [x, y elements of R] Z+ = {1, 2, 3, …} Property Exponents Exponentials Exponents Exponentials Result 1st bx + y = bx · by 23 + 2 = 23 · 22 = 32 2nd bx -y = bx ———- by 23 -2 = 23 = 22 2 3rd bx *y = (bx)y 23 *2 = (23)2 = 64 4th n Z+ bx : n = n√bx 24 : 2 = 2√24 = 4 5th (b · c)x = bx · cx (2 · 3)2 = 22 · 32 = 36 6th (_b_)x = c bx ———- cx (_4_)3 = 2 43 ———- = 23 8 Enzo Exposyto 10
  • 11. Exponentials - 6 Properties - Proofs/Examples PROOFS / EXAMPLES [b, c > 0 and b, c < > 1] 1st b3 · b2 = (b · b · b) · (b · b) = b · b · b · b · b = b5 = b3 + 2 2nd b3 = (b · b · b) = b = b1 = b3 - 2 b2 (b · b) 3rd (b3)2 = (b · b · b) · (b · b · b) = b · b · b · b · b · b = b6 = b3 * 2 4th 2√b4 = 2√(b · b · b · b) = b · b = b2 = b4 : 2 5th b3 · c3 = (b · b · b) · (c · c · c) = b · c · b · c · b · c = … = (b·c)3 6th b3 (b · b · b) b b b b ——- = ————— = —— . —— . —— = … = (—)3 c3 (c · c · c) c c c c Enzo Exposyto 11
  • 12. Exponentials - 2nd property - b superscript 0 EXPONENTIALS - THEIR PROPERTIES - 3 Reference Property Notice Proof Example 2nd Property b0 = 1 b > 0 b ≠ 1 x R b0 = bx - x 20 = 23 - 3 = bx bx = 23 23 = 1 = 1 1 = b0 b > 0 b ≠ 1 1 = 20 Enzo Exposyto 12
  • 13. Exponentials - 2nd property - b superscript (-x) EXPONENTIALS - THEIR PROPERTIES - 4 Reference Property Notice Proof Example 2nd Property b-x = 1 bx b > 0 b ≠ 1 x R b-x = b0-x 2-3 = 20-3 = b0 bx = 20 23 = 1 bx = 1 23 1 = b-x bx b > 0 b ≠ 1 x R 1 = 2-3 23 Enzo Exposyto 13
  • 14. Exponentials - 2nd property - Reciprocal of bx EXPONENTIALS and THEIR PROPERTIES - 5 Reference Property Notice Proof Example 2nd Property bx = 1 b-x b > 0 b ≠ 1 x R 23 = 1 2-3 1 = bx b-x b > 0 b ≠ 1 x R 1 = 23 2-3 Enzo Exposyto 14
  • 16. EXPONENTIAL BASE 2: 23 = 2 × 2 x 2 = 8 The LOGARITHM BASE 2 OF 8 goes the OTHER WAY: Enzo Exposyto 16
  • 17. The logarithm base 2 of 8 is 3, BECAUSE 2 cubed is 8; so the logarithm base 2 of 8 is 3: Enzo Exposyto 17
  • 18. THE LOGARITHM BASE 2 OF 8 IS THE OPERATION THAT ALLOWS US OF GOING BACK TO THE EXPONENT 3 EXPONENTIAL 2x and LOGARITHM BASE 2 EXPONENT x 2x 1 2 2 4 3 8 4 16 Enzo Exposyto 18
  • 19. In other words … THE LOGARITHM BASE 2 OF 8 IS THE EXPONENT 3 … MORE PRECISELY … THE LOGARITHM "3" IS THE EXPONENT WHICH WE HAVE TO PUT ON THE BASE “2” TO GET “8” Enzo Exposyto 19
  • 20. Now, since log2(8) = 3 and 8 = 23 then log2(23) = 3 We can see that THE LOGARITHM "3" IS THE EXPONENT WHICH WE HAVE TO PUT ON THE BASE “2” TO GET “23”
 Enzo Exposyto 20
  • 21. EXPONENTIAL and LOGARITHM with the same base CANCEL EACH OTHER. This is true because exponential and logarithm with the same base are INVERSE OPERATIONS It is just like Addition and Subtraction, Multiplication and Division, Exponentiation and Root, … when they're INVERSE OPERATIONS Enzo Exposyto 21
  • 22. Now, we can introduce the ANTILOGARITHM BASE b: antilogb(x) = bx It's, simply, an EXPONENTIAL and represents the antilogarithm when we operate with a logarithm … It’s such that logb(antilogb(x)) = x The meaning is logb(bx) = x Enzo Exposyto 22
  • 23. And, of course, antilogb(logb(y)) = y The meaning is blogb(y) = y Enzo Exposyto 23
  • 24. These phrases - with 8 - are equivalent log2(8) = 3 < = > 23 = 8 OR 23 = 8 < = > log2(8) = 3 These phrases - with 23 - are equivalent log2(23) = 3 < = > 23 = 23 OR 23 = 23 < = > log2(23) = 3 Enzo Exposyto 24
  • 26. LOGARITHM - DEFINITION DEFINITION Example DEFINITION logb(y) = x iff (if and only if) bx = y log2(8) = 3 because 23 = 8 b > 0 and b < > 1 y > 0 x element of R 2 > 0 and 2 < > 1 8 > 0 3 element of R The logarithm "x" is the exponent which we have to put on the base “b” to get “y” The logarithm "3" is the exponent which we have to put on the base “2” to get “8” Enzo Exposyto 26
  • 27. iff bx = y
Enzo Exposyto 27
  • 28. LOGARITHMS - EXAMPLES - 1 log DEFINITION because log3(9) = 2 The logarithm "2" is the exponent which we have to put on the base “3” to get “9” log3(9) = 2 because 32 = 9 log3(27) = 3 The logarithm "3" is the exponent which we have to put on the base “3” to get “27” log3(27) = 3 because 33 = 27 log4(4) = 1 The logarithm "1" is the exponent which we have to put on the base “4” to get “(4)” log4(4) = 1 because 41 = (4) Enzo Exposyto 28
  • 29. LOGARITHMS - EXAMPLES - 2 log DEFINITION because log4(16) = 2 The logarithm "2" is the exponent which we have to put on the base “4” to get “16” log4(16) = 2 because 42 = 16 log5(25) = 2 The logarithm "2" is the exponent which we have to put on the base “5” to get “25” log5(25) = 2 because 52 = 25 log10(100) = 2 The logarithm "2" is the exponent which we have to put on the base “10” to get “100” log10(100) = 2 because 102 = 100 Enzo Exposyto 29
  • 30. LOGARITHMS - EXAMPLES - 3 log DEFINITION because log4(1) = -1 4 The logarithm "-1" is the exponent which we have to put on the base “4” to get “1” 4 log4(1) = -1 4 because 4-1 = 1 4 log10( 1 ) = -1 10 The logarithm "-1" is the exponent which we have to put on the base “10” to get “ 1 ” 10 log10( 1 ) = -1 10 because 10-1 = 1 10 Enzo Exposyto 30
  • 31. LOGARITHMS - EXAMPLES - 4 log DEFINITION because log1/2(2) = -1 The logarithm "-1" is the exponent which we have to put on the base “1” 2 to get “2” log1/2(2) = -1 because (1)-1 = 2 2 log1/4(4) = -1 The logarithm "-1" is the exponent which we have to put on the base “1” 4 to get “4” log1/4(4) = -1 because (1)-1 = 4 4 Enzo Exposyto 31
  • 32. LOGARITHMS - EXAMPLES - 5 log DEFINITION because log1/2(4) = -2 The logarithm "-2" is the exponent which we have to put on the base “1” 2 to get “4” log1/2(4) = -2 because (1)-2 = 4 2 log1/3(9) = -2 The logarithm "-2" is the exponent which we have to put on the base “1” 3 to get “9” log1/3(9) = -2 because (1)-2 = 9 3 Enzo Exposyto 32
  • 33. LOGARITHMS - EXAMPLES - 6 e = 2.718281828… log DEFINITION because loge(e) = 1 The logarithm "1" is the exponent which we have to put on the base “e” to get “(e)” loge(e) = 1 because e1 = (e) loge(1) = 0 The logarithm "0" is the exponent which we have to put on the base “e” to get “1” loge(1) = 0 because e0 = 1 loge(e2) = 2 The logarithm "2" is the exponent which we have to put on the base “e” to get “(e2)” loge(e2) = 2 because e2 = (e2) Enzo Exposyto 33
  • 35. LOGARITHMS - THEIR PROPERTIES - 1 Name Property Example base = 0 log0(y) is undefined log0(2) is undefined y > 0 For any x R-{0}, 0x ≠ 2 (really 0x = 0) and, then, log0(2) Does Not Exist base = 1 log1(y) is undefined log1(2) is undefined y > 0 For any x R, 1x ≠ 2 (really 1x = 1) and, then, log1(2) Does Not Exist logb(0) logb(0) is undefined log2(0) is undefined b > 0 and b < > 1 For any x R, 2x ≠ 0 (really 2x > 0) and, then, log2(0) Does Not Exist Enzo Exposyto 35
  • 36. LOGARITHMS - THEIR PROPERTIES - 2 Name Property Examples logb(1) logb(1) = 0 log2(1) = 0 because 20 = 1b > 0 and b < > 1 logb(b) logb(b) = 1 log4(4) = 1 because 41 = (4)b > 0 and b < > 1 Enzo Exposyto 36
  • 37. LOGARITHMS - THEIR PROPERTIES - 3 Name Property Example logb(bx) logb(bx) = x log2(23) = 3 because 23 = (23) b > 0 and b < > 1 x element of R blogb(y) blogb(y) = y 2log2(8) = 8 because 23 = 8 b > 0 and b < > 1 y > 0 Enzo Exposyto 37
  • 38. LOGARITHMS - THEIR PROPERTIES - 4 Name Property Examples log of a Power logb(yz) = z logb(y) log2(42) = 2 log2(4) b > 0 and b < > 1 y > 0 z element of R b = 2 y = 4 z = 2 log of a Reciprocal logb(1) = logb(y-1) = - logb(y) y log2(1) = log2(4-1) = - log2(4) 4 b > 0 and b < > 1 y > 0 b = 2 y = 4 Enzo Exposyto 38
  • 39. LOGARITHMS - THEIR PROPERTIES - 5 Name Property Examples log of a Root - 1 logb(n√y) = logb(y) n log2(3√8) = log2(8) 3 b > 0 and b < > 1 y > 0 n Z+ Z+ = {1, 2, 3, …} b = 2 y = 8 n = 3 log of a Root - 2 logb(n√yz) = z logb(y) n log2(3√26) = 6 log2(2) 3 b > 0 and b < > 1 y > 0 z element of R n Z+ Z+ = {1, 2, 3, …} b = 2 y = 2 z = 6 n = 3 Enzo Exposyto 39
  • 40. LOGARITHMS - THEIR PROPERTIES - 6 Name Property Examples log of a Product - 1 logb(y z) = logb(y) + logb(z) log2(4 2) = log2(4) + log2(2) b > 0 and b < > 1 y, z > 0 b = 2 y = 4; z = 2 log of a Product - 2 logb(yn . zp) = n.logb(y)+p.logb(z) log2(43 . 24) = 3.log2(4)+4.log2(2) b > 0 and b < > 1 y, z > 0 n, p elements of R b = 2 y = 4; z = 2 n = 3; p = 4 Enzo Exposyto 40
  • 41. LOGARITHMS - THEIR PROPERTIES - 7 Name Property Examples log of a Quotient - 1 logb(y) = logb(y.z-1) = logb(y)-logb(z) z log2(8) = log2(8.4-1) = log2(8) - log2(4) 4 b > 0 and b < > 1 y, z > 0 b = 2 y = 8; z = 4 log of a Quotient - 2 logb(y) = logb(y) - logb(z) z log2(4) = log2(4) - log2(2) 2 b > 0 and b < > 1 y, z > 0 b = 2 y = 4; z = 2 log of a Quotient - 3 logb(yn) = n.logb(y)-p.logb(z) zp log2(43) = 3.log2(4)-4.log2(2) 24 b > 0 and b < > 1 y, z > 0 n, p elements of R b = 2 y = 4; z = 2 n = 3; p = 4 Enzo Exposyto 41
  • 42. LOGARITHMS - THEIR PROPERTIES - 8 Name Property Examples Base Change - 1 logb(y) = logc(y) logc(b) log2(16) = log4(16) = 2 = 4 log4(2) 1 2b, c > 0 and b, c < > 1 y > 0 Base Switch - 1 logb(c) = logc(c) = 1 logc(b) logc(b) log2(4) = log4(4) = 1 log4(2) log4(2) logc(c) = 1 log4(4) = 1 b, c > 0 and b, c < > 1 b = 2; c= 4 Base Switch - 2 logb(c) * logc(b) = 1 log2(4) * log4(2) = 1 Enzo Exposyto 42
  • 43. LOGARITHMS - THEIR PROPERTIES - 9 Name Property Examples Base Change - 2 logbn(y) = logb(y) n log2-3(8) = log2(8) -3 b > 0 and b < > 1 n element of R, n < > 0 y > 0 b = 2 n = -3 y = 8 Base Change - 3a n . logbn(y) = logb(y) -3 . log2-3(8) = log2(8) n element of R, n < > 0 b > 0 and b < > 1 y > 0 n = -3 b = 2 y = 8 Base Change - 3b logb(y) = n . logbn(y) log2(4) = 2 . log22(4) b > 0 and b < > 1 y > 0 n element of R, n < > 0 b = 2 y = 4 n = 2 Enzo Exposyto 43
  • 44. LOGARITHMS - THEIR PROPERTIES - 10 Name Property Examples Base Change - 4 log1/b(y) = - logb(y) log1/8(8) = - log8(8) b > 0 and b < > 1 y > 0 b = 8 y = 8 Base Change - 5 logb(1) = log1/b(y) y log2(1) = log1/2(4) 4 b > 0 and b < > 1 y > 0 b = 2 y = 4 Enzo Exposyto 44
  • 45. LOGARITHMS - THEIR PROPERTIES - 11 Name Property Examples zlogb(y) zlogb(y) = ylogb(z) 2log2(8) = 8log2(2) z > 0 b > 0 and b < > 1 y > 0 z = 2 b = 2 y = 8 Enzo Exposyto 45
  • 47. log(y) AND ln(y) - THEIR PROPERTIES REMARKS: • log(y) always refers to log base 10, i. e., log(y) = log10(y) Therefore, log(y) = x if and only if 10x = y Enzo Exposyto 47
  • 48. • ln(y) is called the natural logarithm and is used to represent loge(y), where the irrational number e 2.718281828: ln(y) = loge(y) Therefore, ln(y) = x if and only if ex = y Enzo Exposyto 48
  • 49. • Most calculators can directly compute logs base 10 and/or the natural log. For any other base it is necessary to use the change of the base formula: logb(y) = log10(y) = log(y) log2(8) = log(8) log10(b) log(b). log(2) or logb(y) = ln(y) log2(8) = ln(8) ln(b) ln(2)
 Enzo Exposyto 49
  • 50. log(y) AND ln(y) - THEIR PROPERTIES - 1 Property log ln base = 0 base = 10 base = e base = 1 base = 10 base = e logb(0) log(0) is undefined ln(0) is undefined For any x R, 10x ≠ 0 (really 10x > 0) and, then, log(0) Does Not Exist For any x R, ex ≠ 0 (really ex > 0) and, then, ln(0) Does Not Exist Enzo Exposyto 50
  • 51. log(x) AND ln(x) - THEIR PROPERTIES - 2 Property log ln logb(1) log(1) = 0 ln(1) = 0 because 100 = 1 because e0 = 1 logb(b) log(10) = 1 ln(e) = 1 because 101 = 10 because e1 = e Enzo Exposyto 51
  • 52. log(y) AND ln(y) - THEIR PROPERTIES - 3 Property log ln logb(bx) log(10x) = x ln(ex) = x x element of R x R blogb(y) 10log(y) = y eln(y) = y y > 0 y > 0 Enzo Exposyto 52
  • 53. log(y) AND ln(y) - THEIR PROPERTIES - 4 Property log ln log of a Power log(yz) = z log(y) ln(yz) = z ln(y) y > 0 z element of R y > 0 z element of R log of a Reciprocal log(1) = log(y-1) = - log(y) y ln(1) = ln(y-1) = - ln(y) y y > 0 y > 0 Enzo Exposyto 53
  • 54. log(y) AND ln(y) - THEIR PROPERTIES - 5 Property log ln log of a Root - 1 log(n√y) = log(y) n ln(n√y) = ln(y) n n Z+ Z+ = {1, 2, 3, …} y > 0 n Z+ Z+ = {1, 2, 3, …} y > 0 log of a Root - 2 log(n√yz) = z log(y) n ln(n√yz) = z ln(y) n n element of Z+ Z+ = {1, 2, 3, …} y > 0 z element of R n element of Z+ Z+ = {1, 2, 3, …} y > 0 z element of R Enzo Exposyto 54
  • 55. log(y) AND ln(y) - THEIR PROPERTIES - 6 Property log ln log of a Product - 1 log(y z) = log(y) + log(z) ln(y z) = ln(y) + ln(z) y, z > 0 y, z > 0 log of a Product - 2 log(yn . zp) = n.log(y)+p.log(z) ln(yn . zp) = n.ln(y)+p.ln(z) y, z > 0 n, p elements of R y, z > 0 n, p elements of R Enzo Exposyto 55
  • 56. log(y) AND ln(y) - THEIR PROPERTIES - 7 Property log ln log of a Quotient - 1 log(y) = log(y.z-1) = log(y)-log(z) z ln(y) = ln(y.z-1) = ln(y)-ln(z) z y, z > 0 y, z > 0 log of a Quotient - 2 log(y) = log(y) - log(z) z ln(y) = ln(y) - ln(z) z y, z > 0 y, z > 0 log of a Quotient - 3 log(yn) = n.log(y)-p.log(z) zp ln(yn) = n.ln(y)-p.ln(z) zp y, z > 0 n, p elements of R y, z > 0 n, p elements of R Enzo Exposyto 56
  • 57. log(y) AND ln(y) - THEIR PROPERTIES - 8 Property log ln Base Change - 1 log(y) = logc(y) logc(10) ln(y) = logc(y) logc(e) y > 0 c > 0 and c < > 1 y > 0 c > 0 and c < > 1 Base Switch - 1 log(c) = logc(c) = 1 logc(10) logc(10) ln(c) = logc(c) = 1 logc(e) logc(e) logc(c) = 1 logc(c) = 1 c > 0 and c < > 1 c > 0 and c < > 1 Base Switch - 2 log(c) * logc(10) = 1 ln(c) * logc(e) = 1 c > 0 and c < > 1 c > 0 and c < > 1 Enzo Exposyto 57
  • 58. log10(y) AND loge(y) - THEIR PROPERTIES - 9 Property log10 loge Base Change - 2 log10n(y) = log10(y) n logen(y) = loge(y) n n element of R, n < > 0 y > 0 n element of R, n < > 0 y > 0 Base Change - 3a n . log10n(y) = log10(y) n . logen(y) = loge(y) n element of R, n < > 0 y > 0 n element of R, n < > 0 y > 0 Base Change - 3b log10(y) = n . log10n(y) loge(y) = n . logen(y) y > 0 n element of R, n < > 0 y > 0 n element of R, n < > 0 Enzo Exposyto 58
  • 59. log10(y) AND loge(y) - THEIR PROPERTIES - 10 Property log10 loge Base Change - 4 log1/10(y) = - log10(y) log1/e(y) = - loge(y) y > 0 y > 0 Base Change - 5 log10(1) = log1/10(y) y loge(1) = log1/e(y) y y > 0 y > 0 Enzo Exposyto 59
  • 60. log10(y) AND loge(y) - THEIR PROPERTIES - 11 Property log10 loge zlogb(y) zlog10(y) = ylog10(z) zloge(y) = yloge(z) z > 0 y > 0 z > 0 y > 0 Enzo Exposyto 60
  • 62. logb(bx) = x a) The log of bx is the exponent which we have to put on the base b to get bx itself and, therefore, it’s “x” b) If, from x, we ‘go’ to bx and, then, from bx, we ‘go’ to logb(bx), since logb(bx) is the inverse operation of bx, we go back to x … therefore, logb(bx) = x In other words (remembering that bx = antilogb(x)): logb(antilogb(x)) = x 
 Enzo Exposyto 62
  • 63. logb(bx) = x c) Let's set bx = y and, then, we ‘do’ the logarithms in base b of both sides; we get logb(bx) = logb(y) Since bx = y <=> logb(y) = x we can write logb(bx) = logb(y) = x Therefore, logb(bx) = x Q.E.D. Enzo Exposyto 63
  • 65. blogb(y) = y a) If from y, we ‘go’ to logb(y) and, then, from logb(y), we ‘go’ to blogb(y), since blogb(y) is the inverse operation of logb(y), we go back to y … therefore, blogb(y) = y In other words (remembering that blogb(y) = antilogb(logb(y))): antilogb(logb(y)) = y 
 Enzo Exposyto 65
  • 66. blogb(y) = y b) Let's set bx = y and, then, we ‘do’ the log in base b of both sides; we get logb(bx) = logb(y) Remembering that (pages 62-63) logb(bx) = x we can write logb(bx) = logb(y) x = logb(y) or logb(y) = x Now, we ‘do’ the exponentials in base b of both sides and we get blogb(y) = bx Since bx = y we get blogb(y) = y Q.E.D. Enzo Exposyto 66
  • 67. blogb(y) = y c) Let's set logb(y) = x and, then, on the left hand side of the equation, we get blogb(y) = bx Since logb(y) = x <=> bx = y it's blogb(y) = bx = y and, therefore, blogb(y) = y Q.E.D. Enzo Exposyto 67
  • 69. logb(yz) = z . logb(y) 1) Let's set logb(y) = l and, then, the right hand side of the equation becomes z . logb(y) = z . l 2) Since logb(y) = l <=> bl = y or y = bl the left hand side of the equation becomes logb(yz) = logb((bl)z) = logb(blz) Remembering that (pages 62-63) logb(bx) = x it's logb(blz) = lz = z . l and we can write logb(yz) = logb(blz) = z . l 3) Since the left hand side and the right hand side are equal to z . l, they are equal: logb(yz) = z . logb(y) Q.E.D. Enzo Exposyto 69
  • 70. logb(1) = - logb(y) y OR - logb(y) = logb(1) y Remembering that logb(yz) = z . logb(y) [log of a Power, previous page] we get logb(1) = logb(y-1) y = (-1) . logb(y) = - logb(y) Q.E.D.
 Enzo Exposyto 70
  • 72. logb(n√y) = logb(y) n Remembering that logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(n√y) = logb(y1/n) = 1 logb(y) n = logb(y) n Q.E.D.
 Enzo Exposyto 72
  • 73. logb(n√yz) = z . logb(y) n Remembering that logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(n√yz) = logb(yz/n) = z logb(y) n = z . logb(y) n Q.E.D.
 Enzo Exposyto 73
  • 75. logb(y . z) = logb(y) + logb(z) 1) Let's set logb(y) = l logb(z) = m and, then, the right hand side of the equation becomes logb(y) + logb(z) = l + m 2) Since logb(y) = l <=> bl = y or y = bl logb(z) = m <=> bm = z or z = bm the left hand side of the equation becomes logb(y . z) = logb(bl . bm) = logb(bl+m) Remembering that (pages 62-63) logb(bx) = x it's logb(bl+m) = l + m and we can write logb(y . z) = logb(bl+m) = l + m 3) Since the left hand side and the right hand side are equal to l + m, they are equal: logb(y . z) = logb(y) + logb(z) Q.E.D. Enzo Exposyto 75
  • 76. logb(yn . zp) = n . logb(y) + p . logb(z) Remembering that logb(y . z) = logb(y) + logb(z) [previous page] and logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(yn . zp) = logb(yn) + logb(zp) = n . logb(y) + p . logb(z) Q.E.D.
 Enzo Exposyto 76
  • 78. logb(y) = logb(y) - logb(z) z Remembering that logb(y . z) = logb(y) + logb(z) [page 75] and logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(y) = logb(y . 1) = logb(y . z-1) z z = logb(y) + logb(z-1) = logb(y) + (-1) . logb(z) = logb(y) - logb(z) Q.E.D. Enzo Exposyto 78
  • 79. logb(y) = logb(y) - logb(z) z 1) Let's set logb(y) = l logb(z) = m and, then, the right hand side of the equation becomes logb(y) - logb(z) = l - m 2) Since logb(y) = l <=> bl = y or y = bl logb(z) = m <=> bm = z or z = bm the left hand side of the equation becomes logb(y) = logb(bl ) = logb(bl-m) z bm Remembering that (pages 62-63) logb(bx) = x it's logb(bl-m) = l - m and we can write logb(y) = logb(bl-m) = l - m z Enzo Exposyto 79
  • 80. logb(y) = logb(y) - logb(z) z 3) Since the left hand side and the right hand side are equal to l - m, they are equal: logb(y) = logb(y) - logb(z) z Q.E.D. Enzo Exposyto 80
  • 81. logb(yn) = n . logb(y) - p . logb(z) zp Remembering that logb(y) = logb(y) - logb(z) [previous page] z and logb(yz) = z . logb(y) [log of a Power, page 69] we get logb(yn) = logb(yn) - logb(zp) zp = n . logb(y) - p . logb(z) Q.E.D.
 Enzo Exposyto 81
  • 83. logb(y) = logc(y) logc(b) 1) On the left hand side of the equation, let's set logb(y) = x 2) On the right hand side of the equation, since logb(y) = x <=> bx = y or y = bx we substitute y by bx logc(y) = logc(bx) = x . logc(b) [log of a Power, page 69] and we get logc(y) = x . logc(b) = x logc(b) logc(b) 3) Since the left hand side and the right hand side are equal to x, they are equal: logb(y) = logc(y) logc(b) Q.E.D. Enzo Exposyto 83
  • 84. logb(c) = 1 logc(b) It's logb(c) = logc(c) logc(b) Since logc(c) = 1 we get logb(c) = logc(c) logc(b) = 1 logc(b) and, then, logb(c) = 1 logc(b) Q.E.D. Enzo Exposyto 84
  • 85. logb(c) . logc(b) = 1 From logb(c) = 1 logc(b) multiplying both sides by logc(b), it’s logb(c) logc(b) = 1 logc(b) logc(b) and, then, logb(c) logc(b) = 1 Q.E.D. Enzo Exposyto 85
  • 86. logbn(y) = logb(y) n Let's change the base bn by b: logbn(y) = logb(y) logb(bn) [log of a Power, page 69] = logb(y) n logb(b) [remember: logb(b) = 1] = logb(y) n Therefore: logbn(y) = logb(y) n Q.E.D. Enzo Exposyto 86
  • 87. n . logbn(y) = logb(y) OR logb(y) = n . logbn(y) From logbn(y) = logb(y) [previous page] n multiplying both sides by n, it’s n . logbn(y) = logb(y) . n n and, then, n . logbn(y) = logb(y) Q.E.D.
 Enzo Exposyto 87
  • 88. log1/b(y) = - logb(y) a) From logb(y) = n . logbn(y) [previous page] if n = -1 we get logb(y) = (-1) . logb-1(y) [remember: b-1 = 1] b logb(y) = - log1/b(y) [multiplying both sides by (-1)] - logb(y) = log1/b(y) and, then, log1/b(y) = - logb(y) Q.E.D. Enzo Exposyto 88
  • 89. log1/b(y) = - logb(y) b) Let's change the base 1 by b: b log1/b(y) = logb(y) logb(1) [remember: 1 = b-1] b b = logb(y) logb(b-1) [log of a Power, page 69] = logb(y) (-1) logb(b) [remember: logb(b) = 1] = logb(y) (-1) = - logb(y) Q.E.D. Enzo Exposyto 89
  • 90. logb(1) = log1/b(y) y OR log1/b(y) = logb(1) y 1) From page 70: logb(1) = - logb(y) y 2) From previous page: log1/b(y) = - logb(y) 3) Since the first and the second left hand side are equal to - logb(y), they are equal: logb(1) = log1/b(y) y Q.E.D. Enzo Exposyto 90
  • 92. zlogb(y) = ylogb(z) 1) On the left hand side of the equation, let's set logb(y) = x and we get zlogb(y) = zx 2) On the right hand side of the equation, since logb(y) = x <=> bx = y or y = bx we substitute y by bx and we get ylogb(z) = (bx)logb(z) = bxlogb(z) = blogb(zx) [log of a Power, page 69] = zx [blogb(zx) = zx See blogb(y) = y (pages 65-67)] 3) Since the left hand side and the right hand side are equal to zx, they are equal: zlogb(y) = ylogb(z) Q.E.D. 
 Enzo Exposyto 92