log₅ ( 8 ) = log ( 8 ) / log ( 5 ) = 1.292030

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Exercise 3.4.4

log₃ ( x² ) = log₃ ( 3x + 2 ), so x² = 3x + 2. Thus x^{2 -} 3x - 2 = 0. This expression does not factor, so we use the quadratic formula to solve for x: .

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Exercise 3.4.3

e^2x = e^{6 - x} so 2x = 6 - x. Thus, x = 2.

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Exercise 3.4.2

(a) 3^{1 - x} = 3². so 1 - x = 2. Thus x = -1.

(b) log₂ ( x² - x ) = log₂ ( 2 ). So x² - x = 2. That is, x² - x - 2 = 0. So x = 2 or x = -1.

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Exercise 3.4.1

(a) log₃ ( 9 ) = 1 - x, so 2 = 1 - x. Thus, x = 1.

(b) The equivalent statement in exponential form is 2¹ = x² - x. Thus, x² - x - 2 = 0. This can be solved by factoring: ( x - 2 ) ( x + 1 ) = 0, thus x = 2 or x = -1.

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Exercise 3.3.10

(a) log ( 0.01 ) = log₁₀ ( 10^-2 ) = - 2

(b) ln ( e³ ) = log_e ( e³ ) = 3

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Exercise 3.3.9

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Exercise 3.3.8

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Exercise 3.3.7

(a) log_b ( 6 ) = log_b ( 2 ) + log_b ( 6 ) = 1.7075 + 2.7075 = 4.4190

(b) log_b ( 72 ) = log_b ( 2 ^. 6² ) = log_b ( 2 ) + 2 log_b ( 6 ) = 10.5475

(c) log_b ( 1.5 ) = log_b ( 3 / 2 ) = log_b ( 3 ) - log_b ( 2 ) = 1.0000

(d) Since log_b ( 1.5 ) = 1, then b¹ = 1.5. So b = 1.5

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Exercise 3.3.6

b^o = 1 is equivalent to log_b 1 = 0. b¹ = b is equivalent to log_b b = 1.

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Exercise 3.3.5

Let m = log_b ( p ) and n = logb ( q ). Then b^m = p and bⁿ = q, so p / q = b^{m - n}, which is equivalent to m - n = log_b ( p / q ). Replacing m and n yields
log_b ( p ) + logb ( q ) = log_b ( p / q ).

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Exercise 3.3.4

, 

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Exercise 3.3.3

(a) x = log₅( 25 ) is equivalent to 5^x = 25 = 5², so x = 2.

(b) x = log₅( 5 ) is equivalent to 5^x = 5 = 5¹, so x = 1.

(c) x = log₅( 1 / 5 ) is equivalent to 5^x = 1 / 5 = 5^-1, so x = - 1.

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Exercise 3.3.2

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Exercise 3.3.1

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Exercise 3.2.3

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Exercise 3.2.2

A = P e ^{r t}

P = $ 1000, r = 0.15, A = 1000 e^0.45= $1568.31

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Exercise 3.2.1

b = ( 1 + 1 / n )ⁿ

n = 1, b = 2

n = 10, b = ( 1.1 )¹⁰ = 2.593742

n = 100, b = ( 1.01 )¹⁰⁰ = 2.704814

n = 1000, b = ( 1.001 )¹⁰⁰⁰ = 2.716924

n = 1000000, b = ( 1.000001 )^1000000 = 2.718280

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Exercise 3.1.6

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Exercise 3.1.5

R = R_o ( 0.5 )^{t / h =} R_o ( 0.5 )^{6 / 10} = 6.60 grams

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Exercise 3.1.4

n = 12, b = 1.01, t = 3, P = 1000

So A = 1000 ( 1.01 )³⁶ = $1430.77

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Exercise 3.1.3

Compounded monthly: n = 12, b = 1 + 0.12 / 12 = 1.01

( 1.01 )¹² = 1.1268, so the equivalent amount of simple interest is 12.68%.

Compounded daily: n = 365, b = 1 + 0.12 / 365 = 1.0003287671233

( 1.0003287671233 )³⁶⁵ = 1.1275, so the equivalent amount of simple interest is 12.75%

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Exercise 3.1.2

r = 0.25, k = 4 , p = $100

A = 100 ( 1 + 0.25 )⁴ = $244.14

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3.1.1

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