Solve the equation 5x 23x = for the unknown x 4 7x in Software

Encoder QR Code 2d barcode in Software Solve the equation 5x 23x = for the unknown x 4 7x

Solve the equation 5x 23x = for the unknown x 4 7x
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CHAPTER 6 Transcendental Functions
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SOLUTION We take the natural logarithm of both sides: ln(5x 23x ) = ln 4 7x
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Applying the rules for logarithms we obtain ln(5x ) + ln(23x ) = ln 4 ln(7x ) or x ln 5 + 3x ln 2 = ln 4 x ln 7 Gathering together all the terms involving x yields x [ln 5 + 3 ln 2 + ln 7] = ln 4 or x [ln(5 23 7)] = ln 4 Solving for x gives x= EXAMPLE 623
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Simplify the expression B= 5 log7 3 (1/4) log7 16 3 log7 5 + (1/5) log7 32
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ln 4 = log280 4 ln 280
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SOLUTION The numerator of B equals log7 (35 ) log7 (161/4 ) = log7 243 log7 2 = log7 (243/2) Similarly, the denominator can be rewritten as log7 53 + log7 (321/5 ) = log7 125 + log7 2 = log7 (125 2) = log7 250 Putting these two results together, we nd that B= You Try It: What does 3 EXAMPLE 624
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Simplify the expression (log4 9) (log9 16)
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log7 243/2 = log250 (243/2) log7 250 mean (in terms of the natural logarithm function)
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SOLUTION We have
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Transcendental Functions
(log4 9) (log9 15) =
1 log9 4
log9 16 = log4 16 = 2
Calculus with Logs and Exponentials to Arbitrary Bases
641 DIFFERENTIATION AND INTEGRATION OF loga x AND ax
a x dx = ax + C ln a We begin by noting these facts: If a > 0 then d x a = a x ln a; equivalently, (i) dx d 1 (ii) (loga x) = dx x ln a
Math Note: As always, we can state these last formulas more generally as du d u a = au ln a dx dx and 1 du 1 d loga u = dx u dx ln a EXAMPLE 625
Calculate d dx (5x ), d dx (3cos x ), d dx d dx (log4 (x cos x))
(log8 x),
SOLUTION We see that d x (5 ) = 5x ln 5 dx For the second problem, we apply our general formulation with a = 3, u = cos x to obtain d d cos x (3 cos x ln 3 = 3cos x ( sin x) ln 3 ) = 3cos x dx dx
CHAPTER 6 Transcendental Functions
Similarly, d 1 (log8 x) = dx x ln 8 d 1 d (x cos x) (log4 (x cos x)) = (x cos x) ln 4 dx dx = EXAMPLE 626
Integrate 3cot x ( csc2 x) dx
cos x + (x ( sin x)) (x cos x) ln 4
SOLUTION For clarity we set (x) = cot x, (x) = csc2 x Then our integral becomes 3 (x) (x) dx = = 1 ln 3 1 ln 3 3 (x) (x) ln 3 dx
3 (x) + C
Resubstituting the expression for (x) now gives that 3cot x ( csc2 x) dx = You Try It: Evaluate (log6 (x 3 )/x) dx You Try It: Calculate the integral x 3x dx Our new ideas about arbitrary exponents and bases now allow us to formulate a general result about derivatives of powers: For any real exponent a we have d a x = a x a 1 dx EXAMPLE 627
Calculate the derivative of x , x
1 3cot x + C ln 3
, x e
SOLUTION We have
Transcendental Functions
d x = x 1 , dx d 3 x = 3 x 3 1 , dx d e x = e x e 1 dx You Try It: Calculate (d/dx)5sin x x Calculate (d/dx)x 4
GRAPHING OF LOGARITHMIC AND EXPONENTIAL FUNCTIONS
If a > 0 and f (x) = loga x, x > 0, then
Using this information, we can sketch the graph of f (x) = loga x If a > 1 then ln a > 0 so that f (x) > 0 and f (x) < 0 The graph of f is exhibited in Fig 66
If 0 < a < 1 then ln a = ln(1/a) < 0 so that f (x) < 0 and f (x) > 0 The graph of f is sketched in Fig 67 Since g(x) = a x is the inverse function to f (x) = loga x, the graph of g is the re ection in the line y = x of the graph of f (Figs 66 and 67) See Figs 68, 69
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