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
#1:
Instructions: Solve each equation.
$$a)\hspace{.2em}16^{2x + 1}\cdot 64=64$$
$$b)\hspace{.2em}\left(\frac{1}{216}\right)^{-2x}\cdot 216^{-3x}=36^{x - 1}$$
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#2:
Instructions: Solve each equation.
$$a)\hspace{.2em}9 \cdot 81^{1 - 3x}=81^{-2x}$$
$$b)\hspace{.2em}4^{-x - 2}\cdot 4=8$$
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#3:
Instructions: Solve each equation.
$$a)\hspace{.2em}243^{2x}\cdot 81^x=243$$
$$b)\hspace{.2em}\left(\frac{1}{25}\right)^{x - 3}\cdot \left(\frac{1}{125}\right)^{2x}=125$$
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#4:
Instructions: Solve each equation.
$$a)\hspace{.2em}81^{-2x - 3}\cdot 81^{-x}=\frac{1}{3}$$
$$b)\hspace{.2em}\frac{1}{8}\cdot 64^{-x}=64$$
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#5:
Instructions: Solve each equation.
$$a)\hspace{.2em}25^{3x}\cdot 125^{-2x}=1$$
$$b)\hspace{.2em}\left(\frac{1}{2}\right)^{-2x}\cdot 4^{-x}=64$$
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Written Solutions:
#1:
Solutions:
$$a)\hspace{.2em}x=-\frac{1}{2}$$
$$b)\hspace{.2em}x=\frac{2}{5}$$
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#2:
Solutions:
$$a)\hspace{.2em}x=\frac{3}{2}$$
$$b)\hspace{.2em}x=-\frac{5}{2}$$
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#3:
Solutions:
$$a)\hspace{.2em}x=\frac{5}{14}$$
$$b)\hspace{.2em}x=\frac{3}{8}$$
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#4:
Solutions:
$$a)\hspace{.2em}x=-\frac{11}{12}$$
$$b)\hspace{.2em}x=-\frac{3}{2}$$
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#5:
Solutions:
a) All Real Numbers
b) No Solution
