Targeted Review Solutions

Name the value of b for $h (x) = 12 {(0.45)}^{- x}$ $h (x) = 12 {(0.45)}^{- x}$ and state if this represents growth, decay, or neither.

${(0.45)}^{- x} = {(\frac{1}{0.45})}^{x} = {(\frac{20}{9})}^{x} b = \frac{20}{9}$

$b = \frac{20}{9}, growth$

Evaluate: $\log_{49} \frac{1}{343}$

$\begin{array}{rcl} \log_{49} \frac{1}{343} & = & x \\ 49^{x} & = & \frac{1}{343} \\ 7^{2 (x)} & = & 7^{- 3} \\ 2 x & = & - 3 \end{array}$ $\begin{array}{rcl} \log_{49} \frac{1}{343} & = & x \\ 49^{x} & = & \frac{1}{343} \\ 7^{2 (x)} & = & 7^{- 3} \\ 2 x & = & - 3 \end{array}$

$x = - \frac{3}{2}$

Use the properties of logs to correct the right side of the equation.
$\log_{b} {(2 x)}^{3} = 3 \log_{b} 2 + \log_{b} x$

$\log_{b} {(2 x)}^{3} = 3 (\log_{b} 2 + \log_{b} x)$

$3 \log_{b} 2 + 3 \log_{b} x$

Note

Parentheses are needed when the log is expanded because both 2 and x are raised to the 3rd power.

Sketch the graph $y = {(\frac{3}{4})}^{- x}$ using technology.

${(\frac{3}{4})}^{- x} = {(\frac{4}{3})}^{x} a = 1, b = \frac{4}{3}, h = 0, k = 1$

Graph the solution to $125^{2 x - 12} > 25^{x}$ on a number line.

$\begin{array}{rcl} 5^{3} (^{2 x - 12}) & > & 5^{2 (x)} \\ 3 (2 x - 12) & > & 2 x \\ 6 x - 36 & > & 2 x \\ \frac{- 36}{- 4} & > & \frac{- 4 x}{- 4} \\ 9 & < & x \\ x & > & 9 \end{array}$

Write the expression $\log_{a} 4 + 2 \log_{a} x - \log_{a} y$ as a single logarithm.

$\log_{a} 4 + \log_{a} x^{2} - \log_{a} y \log_{a} 4 x^{2} - \log_{a} y$

$\log_{a} \frac{4 x^{2}}{y}$

Solve.

$4^{(2 x - 1)} = 32^{x}$ $4^{(2 x - 1)} = 32^{x}$

${(2^{2})}^{2 x - 1} = {(2^{5})}^{x} 2^{(4 x - 2)} = 2^{5 x} 4 x - 2 = 5 x$

x = –2

$\log_{3} (x - 4) = 2$

$3^{2} = x - 4 9 = x - 4$

x = 13

Multiple Choice

Rewrite the logarithmic equation $\log_{36} \frac{1}{216} = - \frac{3}{2}$ as an exponential equation.

$36^{- \frac{3}{2}} = \frac{1}{216}$
$36^{\frac{1}{216}} = - \frac{3}{2}$
${\frac{1}{216}}^{- \frac{2}{3}} = 36$
$216^{\frac{2}{3}} = 36$

$\log_{a} b = x is a^{x} = b$

Note

This option is not a true statement. The argument and the exponent are switched.

B,C) These options are mathematically true but do not represent the given logarithmic equation.

Describe the transformation of the exponential function from f(x) to g(x) when $f (x) = {(3)}^{x} and g (x) = {(3)}^{x - 6} + 4$

g(x) is translated right 4 units and up 6 units from f(x)
g(x) is translated right 6 units and up 4 units from f(x)
g(x) is translated left 4 units and up 6 units from f(x)
g(x) is translated left 6 units and up 4 units from f(x)

f(x) g(x)

a 1 1

b 3 3

h 0 6

k 0 4

Note

A, C) The values of h and k are switched. 

C, D) The translation does not move to the left.

Solve: $\log_{2} (x + 2) + \log_{2} (x + 3) = 2$

1, 2
–0.5
–2, –1
–1.5

$\begin{array}{rcl} \log_{2} (x + 2) (x + 3) & = & 2 \\ \log_{2} (x^{2} + 5 x + 6) & = & 2 \\ x^{2} + 5 x + 6 & = & 2^{2} \\ x^{2} + 3 x + 6 & = & 4 \\ x^{2} + 3 x + 2 & = & 0 \\ (x + 2) (x + 1) & = & 0 \\ x & = & - 2, - 1 \end{array}$

Note

This option includes the factors of 2 but not the solution.
This option is the solution if you incorrectly combined the terms to $2 x + 5 = 2^{2}$
This option is the solution if you incorrectly combined the terms to $2 x + 5 = 2$

In the equation $\frac{3^{5 x}}{3^{7}} = 3^{3}$ , which expression represents the left side of the equation written with all terms in the numerator?

2
$3^{\frac{5}{7} x}$
$3^{5 x + 7}$
$3^{5 x - 7}$

$\frac{3^{5 x}}{3^{7}} = 3^{5 x} \cdot 3^{- 7} = 3^{5 x - 7}$

Note

This option is the value of x, but it does not answer the question.

B, C) These options do not apply the exponent rules correctly.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L37	L39	L40	L37	L38	L40	L38	L40	L39	L37	L40	—

L = Lesson in this level, A1 = Algebra 1: Principles of Secondary Mathematics

	f(x)	g(x)
a	1	1
b	3	3
h	0	6
k	0	4