When working with logarithmic equations, we utilize a variety of techniques to obtain a solution. In some cases, we will need to convert into exponential form to get a solution. In other cases, we will rely on some basic properties of logarithms to obtain our solution.

Test Objectives
• Demonstrate the ability to convert between exponential and logarithmic form
• Demonstrate the ability to Solve Logarithmic Equations
Solving Logarithmic Equations Practice Test:

#1:

Instructions: solve each equation.

$$a)\hspace{.2em}{-}6\text{log}_{9}(-9x-6)-9=-15$$

$$b)\hspace{.2em}2\text{log}_{3}(-6x-5)-9=-13$$

#2:

Instructions: solve each equation.

$$a)\hspace{.2em}9\text{log}_{7}(-6x-5)-10=-19$$

$$b)\hspace{.2em}\text{log}_{6}(-13x-3)=\text{log}_{6}(x^2 + 27)$$

#3:

Instructions: solve each equation.

$$a)\hspace{.2em}\text{log}_{7}(8x + 1)=\text{log}_{7}(x^2 + 1)$$

$$b)\hspace{.2em}\text{log}(3 - 4x) - \text{log}(7)=1$$

#4:

Instructions: solve each equation.

$$a)\hspace{.2em}\text{log}_{3}(3 - 5x) + \text{log}_{3}(7)=\text{log}_{3}(24)$$

$$b)\hspace{.2em}\text{log}_{5}(4x^2 + 7) - \text{log}_{5}(9)=1$$

#5:

Instructions: solve each equation.

$$a)\hspace{.2em}\text{ln}(x - 13) + \text{ln}(x - 3)=\text{ln}(75)$$

$$b)\hspace{.2em}\text{ln}(x + 3) - \text{ln}(x - 1)=\text{ln}(4)$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}x=-\frac{5}{3}$$

$$b)\hspace{.2em}x=-\frac{23}{27}$$

#2:

Solutions:

$$a)\hspace{.2em}x=-\frac{6}{7}$$

$$b)\hspace{.2em}x=-3, -10$$

#3:

Solutions:

$$a)\hspace{.2em}x=8,0$$

$$b)\hspace{.2em}x=-\frac{67}{4}$$

#4:

Solutions:

$$a)\hspace{.2em}x=-\frac{3}{35}$$

$$b)\hspace{.2em}x=\frac{\pm \sqrt{38}}{2}$$

#5:

Solutions:

$$a)\hspace{.2em}x=18$$

$$b)\hspace{.2em}x=\frac{7}{3}$$