Question

what is the inverse of the given relation?

1. $y = 3x + 9$

a. $y = \frac{1}{3}x + 3$ c. $y = 3x + 3$
b. $y = 3x - 3$ d. $y = \frac{1}{3}x - 3$
write the equation in logarithmic form.

1. $125^{\frac{4}{3}} = 625$

a. $\log_{\frac{4}{3}} 625 = 125$ c. $\log_{125} 625 = \frac{4}{3}$
b. $3 \log_{4} 625 = 125$ d. $\log_{625} 125 = \frac{3}{4}$
write the equation in exponential form.

1. $\log_{4} \frac{1}{16} = -2$

a. $4^{\frac{1}{2}} = 16$ c. $16^{\frac{1}{2}} = 4$
b. $4^{2} = 16$ d. $4^{-2} = \frac{1}{16}$
evaluate the logarithm.

1. $\log_{5} \frac{1}{625}$

a. $-5$ b. $5$ c. $-4$ d. $4$
use natural logarithms to solve the equation. round to the nearest thousandth

1. $8e^{4x + 8} = 15$

a. $-0.033$ b. $0.264$ c. $-1.843$ d. $2.157$
expand the logarithmic expression.

1. $\log_{3} 11p^{3}$

a. $\log_{3} 11 \cdot 3 \log_{3} p$ c. $\log_{3} 11 + 3 \log_{3} p$
b. $\log_{3} 11 - 3 \log_{3} p$ d. $11 \log_{3} p^{3}$

Explanation:

Question 71

Step1: Swap x and y

Given \( y = 3x + 9 \), swap \( x \) and \( y \) to get \( x = 3y + 9 \).

Step2: Solve for y

Subtract 9 from both sides: \( x - 9 = 3y \).
Divide by 3: \( y = \frac{1}{3}x - 3 \).

Recall the exponential - logarithmic conversion: \( a^b = c \) is equivalent to \( \log_a c = b \).
Given \( 125^{\frac{4}{3}} = 625 \), here \( a = 125 \), \( b=\frac{4}{3} \), \( c = 625 \).
So the logarithmic form is \( \log_{125}625=\frac{4}{3} \).

Recall the logarithmic - exponential conversion: \( \log_a c = b \) is equivalent to \( a^b = c \).
Given \( \log_4\frac{1}{16}=-2 \), here \( a = 4 \), \( b=-2 \), \( c=\frac{1}{16} \).
So the exponential form is \( 4^{-2}=\frac{1}{16} \).

Answer:

d. \( y = \frac{1}{3}x - 3 \)

Question 72