print barcode image c# A NEW APPROACH TO LOGARITHMS in Software

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A NEW APPROACH TO LOGARITHMS
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When you studied logarithms in the past you learned the formula
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this says that logs convert multiplication to addition It turns out that this property alone uniquely determines the logarithm function Let (x) be a differentiable function with domain the positive real numbers and whose derivative function (x) is continuous Assume that satis es the multiplicative law (x y) = (x) + (y) ( ) for all positive x and y Then it must be that (1) = 0 and there is a constant C such that C (x) = x In other words x C dt (x) = 1 t A function (x) that satis es these properties is called a logarithm function The particular logarithm function which satis es (1) = 1 is called the natural
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log(x y) = log x + log y;
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CHAPTER 6 Transcendental Functions
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logarithm: In other words, natural logarithm = ln x =
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1 dt t
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For 0 < x < 1 the value of ln x is the negative of the actual area between the graph and the x-axis This is so because the limits of integration, x and 1, occur in x reverse order: ln x = 1 (1/t) dt with x < 1
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Fig 61
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Notice the following simple properties of ln x which can be determined from looking at Fig 61: (i) When x > 1, ln x > 0 (after all, ln x is an area) (ii) When x = 1, ln x = 0 (iii) When 0 < x < 1, ln x < 0
since
1 dt = t
1 dt < 0 t
(iv) If 0 < x1 < x2 then ln x1 < ln x2 We already know that the logarithm satis es the multiplicative property By applying this property repeatedly, we obtain that: If x > 0 and n is any integer then ln(x n ) = n ln x A companion result is the division rule: If a and b are positive numbers then a = ln a ln b ln b EXAMPLE 61
Simplify the expression A = ln a 3 b2 c 4 d
Transcendental Functions
SOLUTION We can write A in simpler terms by using the multiplicative and quotient properties: A = ln(a 3 b2 ) ln(c 4 d) = [ln a 3 + ln(b2 )] [ln(c 4 ) + ln d] = [3 ln a + 2 ln b] [( 4) ln c + ln d] = 3 ln a + 2 ln b + 4 ln c ln d The last basic property of the logarithm is the reciprocal law: For any x > 0 we have ln(1/x) = ln x EXAMPLE 62
Express ln(1/7) in terms of ln 7 Express ln(9/5) in terms of ln 3 and ln 5
SOLUTION We calculate that ln(1/7) = ln 7, ln(9/5) = ln 9 ln 5 = ln 32 ln 5 = 2 ln 3 ln 5 You Try It: Simplify ln(a 2 b 3 /c5 )
THE LOGARITHM FUNCTION AND THE DERIVATIVE
Now you will see why our new de nition of logarithm is so convenient If we want to differentiate the logarithm function we can apply the Fundamental Theorem of Calculus: d d ln x = dx dx More generally, d 1 du ln u = dx u dx EXAMPLE 63
Calculate d dx ln(4 + x), d dx ln(x 3 x), d dx ln(cos x), d dx
1 1 dt = t x
[(ln x)5 ],
d dx
[(ln x) (cot x)]
CHAPTER 6 Transcendental Functions
SOLUTION For the rst problem, we let u = 4 + x and du/dx = 1 Therefore we have d 1 d 1 ln(4 + x) = (4 + x) = dx 4 + x dx 4+x Similarly, d 3 1 3x 2 1 d (x x) = 3 ln(x 3 x) = 3 dx x x dx x x d 1 d sin x ln(cos x) = (cos x) = dx cos x dx cos x 5(ln x)4 d d 1 [(ln x)5 ] = 5(ln x)4 (ln x) = 5(ln x)4 = dx dx x x d d d [(ln x) (cot x)] = ln x (cot x) + (ln x) cot x dx dx dx 1 = cot x + (ln x) ( csc2 x) x You Try It: What is the derivative of the function ln(x 3 + x 2 ) Now we examine the graph of y = ln x Since d (ln x) = dx d2 (ii) (ln x) = dx (iii) ln(1) = 0, (i) 1 > 0, x d 1 dx x
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