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.

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

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

#2:

Instructions: solve each equation.

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

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

#3:

Instructions: solve each equation.

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

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

#4:

Instructions: solve each equation.

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

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

#5:

Instructions: solve each equation.

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

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

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:

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

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