Unit 7

Units 1 to 6 Review

Practice 7.2C Solutions

Use these solutions to correct your work. When finished, give yourself a grade using the Practice Assessment rubric.

Page 419 and 420, questions 1 to 3, 5, 6, 7a, 7c, 8, and 11a.

Question 1, page 419

D

Additional Notes

The graph of the inverse will be a reflection of the original across the line

y = x

.

Question 2, page 419

A

Additional Notes

Use logarithmic laws to express as an isolated logarithm.

k = \log_{h} 5^{- 1}

Then, express in exponential form.

\begin{array}{l} h^{k} = 5^{- 1} \\ h^{k} = \frac{1}{5} \end{array}

Question 3, page 419

B

Additional Notes

Compare to

y = a \log_{3} (x - h)

.

\begin{array}{rcl} y & = & \log_{3} \sqrt{(x + 7)} \\ = & \log_{3} {(x + 7)}^{\frac{1}{2}} \\ = & \frac{1}{2} \log_{3} (x + 7) \end{array}

There is a vertical stretch about the

x

-axis by a factor of

\frac{1}{2}

and a horizontal translation

7

units to the left.

Question 5, page 419

C

Additional Notes

Use logarithmic laws to simplify.

\begin{array}{rcl} \log_{2} 8 \sqrt{3} & = & \log_{2} 8 + \log_{2} \sqrt{3} \\ = & \log_{2} {(2)}^{3} + \log_{2} 3^{\frac{1}{2}} \\ = & 3 \log_{2} 2 + \frac{1}{2} \log_{2} 3 \\ = & 3 (1) + \frac{1}{2} \log_{2} 3 \end{array}

Now replace

\log_{2} 3

with

x

.

= 3 + \frac{1}{2} x

Question 6, page 419

B

Additional Notes

Concentration of Hydrogen ion for acetic acid:

\begin{array}{rcl} 2.9 & = & - \log [H^{+}] \\ 2.9 & = & \log {[H^{+}]}^{- 1} \\ 2.9 & = & \log [\frac{1}{H^{+}}] \\ 10^{2.9} & = & [\frac{1}{H^{+}}] \\ H^{+} & = & \frac{1}{10^{2.9}} \end{array}

Formic Acid is

4

times as concentrated as acetic acid.

\begin{array}{rcl} 4 H^{+} & = & concentration of formic acid \\ = & 4 (\frac{1}{10^{2.9}}) \end{array}

\begin{array}{rcl} H_{F}^{+} & = & concentration of formic acid \\ Let H_{F}^{+} & = & 4 (\frac{1}{10^{2.9}}) \\ = & \frac{4}{10^{2.9}} \end{array}

Now, substitute

H_{F}^{+} = \frac{4}{10^{2.9}}

back into the pH equation.

\begin{array}{rcl} p H_{F}^{+} & = & - \log [\frac{4}{10^{2.9}}] \\ ≐ & 2.3 \end{array}

Questions 7a and 7c, page 420

a.

\frac{1}{81}

b.

5

Additional Notes

a.

\begin{array}{rcl} \log_{9} x & = & - 2 \\ 9^{- 2} & = & x \\ \frac{1}{9^{2}} & = & x \\ \frac{1}{81} & = & x \end{array}

b.

\begin{array}{rcl} \log_{3} (\log_{x} 125) & = & 1 \\ 3^{1} & = & \log_{x} 125 \\ 3 & = & \log_{x} 125 \\ x^{3} & = & 125 \\ x & = & \sqrt[3]{125} \\ x & = & 5 \end{array}

Question 8, page 420

m = 2.5, n = 0.5

Additional Notes

Determine two equations in

m

and

n

.

Equation 1

\begin{array}{rcl} 5^{m + n} & = & 125 \\ 5^{m + n} & = & 5^{3} \\ m + n & = & 3 \\ m & = & 3 - n \end{array}

Equation 2

\begin{array}{rcl} \log_{m - n} 8 & = & 3 \\ {(m - n)}^{3} & = & 8 \\ m - n & = & \sqrt[3]{8} \\ m - n & = & 2 \\ m & = & 2 + n \end{array}

From equation 1, we have

m = 3 - n

and from equation 2, we have

m = 2 + n

.

\begin{array}{rcl} 3 - n & = & 2 + n \\ 1 & = & 2 n \\ \frac{1}{2} & = & n \end{array}

And then,

m = 3 - n

or

m = 3 - \frac{1}{2}

; therefore,

m = 2.5

.

Question 11a, page 420

a.

No solution

\begin{array}{rcl} \log_{2} (x - 4) - \log_{2} (x + 2) & = & 4 \\ \log_{2} (\frac{x - 4}{x + 2}) & = & 4 \\ 2^{4} & = & \frac{x - 4}{x + 2} \\ 16 & = & \frac{x - 4}{x + 2} \\ 16 (x + 2) & = & x - 4 \\ 16 x + 32 & = & x - 4 \\ 15 x & = & - 36 \\ x & = & - \frac{36}{15} \end{array}

Recall for

\log x = y, x > 0

. Given

\log_{2} (x - 4)

and

\log_{2} (x + 2)

,

x - 4 > 0

and

x + 2 > 0

x > 4

and

x > - 2

.

To satisfy both conditions,

x > 4

.

The solution value is not greater than

4

. There is no solution.