Exercise 39 Page 260

a By performing inverse operations we can solve for

x .

\frac{2 x - 8}{1 0} = 6

$LHS \cdot 10 = RHS \cdot 10$

2 x - 8 = 60

$LHS + 8 = RHS + 8$

2 x = 68

$LHS / 2 = RHS / 2$

x = 34

b Examining the equation, we see that the power on the left-hand side can be rewritten as a power with base

3 .

If we do that, we can then equate the exponents on the two sides of the equation and solve for

x .

9^{x} = 3^{40}

Write as a power

(3^{2})^{x} = 3^{40}

$(a^{m})^{n} = a^{m \cdot n}$

3^{2 x} = 3^{40}

Equate exponents

2 x = 40

$LHS / 2 = RHS / 2$

x = 20

5.1.1 Representing Exponential Growth p.250-252

5.1.2 Rebound Ratios p.255-256

5.1.3 The Bouncing Ball and Exponential Decay p.259-260