### About Solving Exponential Equations with Like Bases:

In some cases, we will be asked to solve exponential equations with like bases. To do this, we can use a simple rule that tells us: if ay = ax, then x = y (a is greater than 0 and a is not equal to 1). So basically, if we have like bases, we can set the exponents equal to each other and solve the resulting equation.

Test Objectives
• Demonstrate an understanding of the rules of exponents
• Demonstrate the ability to solve exponential equations with like bases
Solving Exponential Equations with Like Bases Practice Test:

#1:

Instructions: Solve each equation.

$$a)\hspace{.2em}16^{2x + 1}\cdot 64=64$$

$$b)\hspace{.2em}\left(\frac{1}{216}\right)^{-2x}\cdot 216^{-3x}=36^{x - 1}$$

#2:

Instructions: Solve each equation.

$$a)\hspace{.2em}9 \cdot 81^{1 - 3x}=81^{-2x}$$

$$b)\hspace{.2em}4^{-x - 2}\cdot 4=8$$

#3:

Instructions: Solve each equation.

$$a)\hspace{.2em}243^{2x}\cdot 81^x=243$$

$$b)\hspace{.2em}\left(\frac{1}{25}\right)^{x - 3}\cdot \left(\frac{1}{125}\right)^{2x}=125$$

#4:

Instructions: Solve each equation.

$$a)\hspace{.2em}81^{-2x - 3}\cdot 81^{-x}=\frac{1}{3}$$

$$b)\hspace{.2em}\frac{1}{8}\cdot 64^{-x}=64$$

#5:

Instructions: Solve each equation.

$$a)\hspace{.2em}25^{3x}\cdot 125^{-2x}=1$$

$$b)\hspace{.2em}\left(\frac{1}{2}\right)^{-2x}\cdot 4^{-x}=64$$

Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}x=-\frac{1}{2}$$

$$b)\hspace{.2em}x=\frac{2}{5}$$

#2:

Solutions:

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

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

#3:

Solutions:

$$a)\hspace{.2em}x=\frac{5}{14}$$

$$b)\hspace{.2em}x=\frac{3}{8}$$

#4:

Solutions:

$$a)\hspace{.2em}x=-\frac{11}{12}$$

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

#5:

Solutions:

a) All Real Numbers

b) No Solution