5 x f x when x 0 f 0 1 so 0 1 is an intercept f x 0

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( ) 5 x f x = When x = 0, f (0) = 1, so (0, 1) is an intercept. f ( x ) = 0 has no solution. 1 lim 5 lim 0 5 x x t t →−∞ →−∞ = so y = 0 is a horizontal asymptote. lim 5 x t →+∞ = +∞ so f ( x ) increases without bound as x increases. 2. 3. 2 ( ) ln f x x = Note that the domain of f is x 0, so x = 0 is a vertical asymptote. When f ( x ) = 0, x = ± 1 so ( 1, 0) and (1, 0) are intercepts. 2 2 lim ln lim , x x x x →−∞ →∞ = = +∞ so f increases without bound as x decreases and as x increases. 4. 5. (a) ( ) kx f x Ae = Since f (0) = 10, 0 10 , Ae = or A = 10 and ( ) 10 . kx f x e = Since f (1) = 25, (1) 25 10 5 2 5 ln ln 2 5 ln 2 k k k e e e k = = = = − So, ( ) ( ) 5 5 2 2 ln ln ( ) 10 10 t t f t e e − − = = 4 4ln(5/2) ln(5/20) 4 (4) 10 10 5 10 2 3125 8 f e e = = = = (b) ( ) kx f x Ae = Since f (1) = 3, (1) 3 , k Ae = or 3 . k A e = Since f (2) = 10, (2) 10 , k Ae = or
Chapter 4. Exponential and Logarithmic Functions 453 2 10 . k A e = So, 2 3 10 3 10 k k k e e e = = and 3 9 3 . 10 10 A = (3) 3 3 9 (3) 10 9 ( ) 10 9 3 10 10 9 1000 10 27 100 3 k k f e e = = = = = (c) ( ) 30 kx f x Ae = + Since f (0) = 50, 0 50 30 , Ae = + or A = 20 and ( ) 30 20 . kx f x e = + Since f (3) = 40, (3) 3 3 40 30 20 10 20 1 2 k k k e e e = + = = (9) 3 3 3 (9) 30 20 30 20( ) 1 30 20 2 5 30 2 65 2 k k f e e = + = + = + = + = (d) 6 ( ) 1 kt f t Ae = + Since f (0) = 3, 0 6 3 , 1 Ae = + or A = 1. Now, 6 ( ) . 1 kt f t e = + Since f (5) = 2, 5 5 (5) 6 2 , 1 3, 2. 1 k k k e e e = + = + So, (10) 5 2 2 6 (10) 1 6 1 ( ) 6 1 (2) 6 5 k k f e e = + = + = + = 6. (a) 5 ln 5 since ln x e x e = (b) ln 2 ln 2 since x e x e = (c) 3 3ln 4 ln 2 ln 4 ln 2 64 ln 2 ln32 32 e e e e = = = = (d) ( ) ( ) ( )( ) 2 2 2 2 ln 9 ln 3 ln 9 3 ln 27 3ln3 e e e e + = = = 7. 0.04 0.04 8 2 4 0.04 ln4 25ln4 x x e e x x = = = = 8. 6 6 5 1 4 4 4 6 ln1 0 0 x x e x x = + = = = = 9. 4 ln x = 8, ln x = 2, or 2 7.389. x e = 10. 3 3 5 ln5 ln ln5 3 3 1.864 ln5 x x e e x x = = = = =
454 Chapter 4. Exponential and Logarithmic Functions 11. 9 log (4 1) 2 x = if and only if 2 4 1 9 41 4 82, or 2 x x x = = 12. ln( 2) 3 ln( 1) x x + = + 3 ln( 1) ln( 2) 1 ln 2 x x x x = + + = 3 3 3 3 3 3 3 1 2 2 1 ( 1) 1 2 1 2 1 x e x e x e x e x e e x e + = =+ =+ + = 13. 2 2 0 x x e e + = Letting , x u e = 2 2 0 ( 2)( 1) 0 or, 2, 1. u u u u u + = + = = − If u = 2, 2 x e = − and there is no solution. If u = 1, 1, x e = so x = 0. 14. Let . = x u e 2 2 2 3 0 2 3 0 ( 3)( 1) 0 + = + = + = x x e e u u u u 3 or 1 3 or 1 = − = = − = x x u u e e x e is never negative, so 3 = − x e is extraneous. Thus, 1 = x e so x = 0 is the only solution. 15. 2 x y x e = 2 ( )( ) ( )(2 ) ( 2) x x x dy x e e x dx xe x = + = + 16. 3 5 2 x y e + = 3 5 3 5 2 (3 5) 6 x x dy d e x e dx dx + + = + 17. 2 ln 2 ln y x x x x = = 1 (2 ) (ln )(2) 2(1 ln ) dy x x x dx x = + = + 18. 2 2 ln 4 1 1 ln( 4 1) 2 y x x x x = + + = + 2 2 2 1 1 2 ( 4 1) 2 4 1 4 1 dy d x x x dx dx x x x x + = + + + + + 19. 2 2 3 ln( ) 2 log ( ) ln ln3 ln3 x y x x = = 2 1 2 ln3 ln3 dy dx x x = =
Chapter 4. Exponential and Logarithmic Functions 455 20.

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