Targeted Review Solutions

Write the equation of the exponential function g(x) that translates f(x) up 3 spaces and left 5 spaces.
$f (x) = 4^{x + 2} - 6$

$h = - 2 - 5 = - 7 k = - 6 + 3 = - 3$

$g (x) = 4^{x + 7} - 3$

Solve: ${(\frac{1}{27})}^{x - 2} = 9^{(- x + 6)}$

${(\frac{1}{3^{3}})}^{x - 2} = {(3^{2})}^{- x + 6} {(3^{- 3})}^{x - 2} = {(3^{2})}^{- x + 6} 3^{(- 3 x + 6)} = 3^{(- 2 x + 12)} - 3 x + 6 = - 2 x + 12$

$x = - 6$

Write in exponential form. Do not solve.
A) $\log_{216} 36$
B) $\ln 7.5$

$\log_{216} 36 = x$
$\ln 7.5 = x$

$216^{x} = 36$
$e^{x} = 7.5$

Evaluate: $\log_{8} 16$

$\begin{array}{rcl} \log_{8} 16 & = & x \\ 8^{x} & = & 16 \\ 2^{3 (x)} & = & 2^{4} \\ 3 x & = & 4 \end{array}$

$\frac{4}{3}$

Solve with common logs: $17^{x} = 254$

$\log 17^{x} = \log 254 x \log 17 = \log 254$

$x = \frac{\log 254}{\log 17}$

Solve: $2 \ln x + \ln 36 = \ln 18 + \ln 8$

$\begin{array}{rcl} \ln x^{2} + \ln 36 & = & \ln (18 \cdot 8) \\ \ln (x^{2} \cdot 36) & = & \ln (144) \\ \frac{36 x^{2}}{36} & = & \frac{144}{36} \\ \sqrt{x^{2}} & = & \pm \sqrt{4} \\ x & = & - 2, 2 \end{array}$

x = 2

Express as a single logarithm.
$\log_{5} x - 2 (\log_{5} 2 + \log_{5} y)$

$\log_{5} x - 2 (\log_{5} 2 y) \log_{5} x - \log_{5} {(2 y)}^{2} \log_{5} x - \log_{5} 4 y^{2}$

$\log_{5} (\frac{x}{4 y^{2}})$

Expand: $\log_{x} (\frac{{(y - 2)}^{3}}{8})$

$\log_{x} {(y - 2)}^{3} - \log_{x} 8$

$3 \log_{x} (y - 2) - \log_{x} 8$

Multiple Choice

Solve: $4^{x + 5} > 1$

Note

A, B) The options simplify the expression incorrectly to $\log 0.25 - 5$ .
B, C) The value of x is greater than, not less than.

$\begin{array}{rcl} \log 4^{x + 5} & > & \log 1 \\ (x + 5) \log 4 & > & \log 1 \\ x + 5 & > & \frac{\log 1}{\log 4} \\ x & > & \frac{\log 1}{\log 4} - 5 \\ x & > & 0 - 5 \\ x & > & - 5 \end{array}$

Name the end behavior of the given function.

$As x \to + \infty, f (x) \to + \infty, and as x \to 0, f (x) \to - \infty$
$As x \to + \infty, f (x) \to - \infty, and as x \to 0, f (x) \to + \infty$
$As x \to + \infty, f (x) \to + \infty, and as x \to - \infty, f (x) \to 0$
$As x \to + \infty, f (x) \to - \infty, and as x \to - \infty, f (x) \to 0$

Domain: all real numbers
Range: y > 0

Note

A, B) The x-values are all real numbers, so x does not approach zero.

When x is increasing, f(x) is also increasing.

Write $\log_{5} 2$ as a common log.

$\frac{\log 2}{\log 5}$
$\frac{\log 5}{\log 2}$
$\frac{\log_{2} 5}{\log_{2} 2}$
$\frac{\log_{5} 2}{\log_{5} 5}$

Note

This option is the reciprocal of the answer.

C, D) The expression has a common base, but does not use the common log.

Change of base for a common logarithm: $\log_{b} X = \frac{\log X}{\log b}$

Write $3 \ln y + \ln 2 - \ln (y - 5)$ as a single logarithm.

$\ln (\frac{2 y^{3}}{y - 5})$
$\ln (\frac{y^{3} + 2}{y - 5})$
$\ln (\frac{6 y}{\frac{y}{5}})$
$\ln (\frac{2 y^{3}}{\frac{y}{5}})$

$\ln y^{3} + \ln 2 - \ln (y - 1) \ln (y^{3} \cdot 2) - \ln (y - 1) \ln (\frac{2 y^{3}}{y - 1})$

Note

B, C) The Product Rule was incorrectly applied in the numerator.
C, D) The Quotient Rule was incorrectly applied in the denominator.

Problem	1	2	3	4	5	6	7	8	9	10	11	12
Origin	L37	L38	L39, 42	L39	L41	L42	L40	L40	L38	L37	L41	L42

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