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# Solving Equations with Exponents

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Practice Solving Equations with Exponents
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# Exponential Equations

The following exponential equation is one in which the variable appears in the exponent.

$9^{x+1}=\sqrt{27}$

How can you solve this type of equation where you can't isolate the variable?

### Guidance

When an equation has exponents, sometimes the variable will be in the exponent and sometimes it won't. There are different strategies for solving each type of equation.

• When the variable is in the exponent: Rewrite each side of the equation so that the bases of the exponent are the same. Then, create a new equation where you set the exponents equal to each other and solve (see Example A).
• When the variable is not in the exponent: Manipulate the equation so the exponent is no longer there (see Example B). Or, rewrite each side of the equation so that both sides have the same exponent. Then, create a new equation where you set the bases equal to each other and solve (see Example C).

#### Example A

Solve the following exponential equation:

$25^{x-3}=\left(\frac{1}{5}\right)^{3x+18}$

Solution: The variable appears in the exponent. Write both sides of the equation as a power of 5.

$({\color{red}5^2})^{x-3}=({\color{red}5^{-1}})^{3x+18}$

Apply the law of exponents for raising a power to a power $\boxed{(a^m)^n=a^{mn}}$ .

$& 5^{{\color{red}2(x-3)}}=5^{{\color{red}-1(3x+18)}} && \text{Simplify the exponents}.\\& 5^{{\color{red}2x-6}}=5^{{\color{red}-3x-18}} && \text{The bases are the same so the exponents are equal quantities}.\\& 2x-6=-3x-18 && \text{Set the exponents equal to each other and solve the equation}.\\& 2x-6 {\color{red}+6}=-3x-18 {\color{red}+6}\\& 2x=-3x {\color{red}-12}\\& 2x {\color{red}+3x}=-3x {\color{red}+3x}-12\\& {\color{red}5x}=-12\\& \frac{5x}{{\color{red}5}}=\frac{-12}{{\color{red}5}}\\& \frac{\cancel{5}x}{\cancel{5}}={\color{red}\frac{-12}{5}}\\& \boxed{x=\frac{-12}{5}}$

#### Example B

Solve the following exponential equation:

$4(x-2)^{\frac{1}{2}}=16$

Solution: The variable appears in the base.

$& 4(x-2)^{\frac{1}{2}}=16 && \text{Divide both sides of the equation by} \ 4.\\& \frac{4(x-2)^{\frac{1}{2}}}{{\color{red}4}}=\frac{16}{{\color{red}4}}\\& \frac{\cancel{4}(x-2)^{\frac{1}{2}}}{\cancel{4}}=\frac{\overset{{\color{red}4}}{\cancel{16}}}{\cancel{4}}\\& (x-2)^{\frac{1}{2}}=4 && \text{Multiply the exponents on each side of the equation by the reciprocal of} \ \frac{1}{2}.\\& \left[(x-2)^{\frac{1}{2}}\right]^{{\color{red}2}} && \text{Apply the law of exponents} \ \boxed{(a^m)^n=a^{mn}}.\\& (x-2)^{{\color{red}\frac{1}{2} \times 2}}=(4)^{{\color{red}1 \times 2}} && \text{Simplify the exponents}.\\& (x-2)^{\color{red}1}={\color{red}4^2}\\& {\color{red}x-2}={\color{red}16} && \text{Solve the equation}.\\& x-2 {\color{red}+2}=16 {\color{red}+2}\\& x={\color{red}18}\\& \boxed{x=18}$

#### Example C

Solve the following exponential equation:

$(2x-4)^{\frac{2}{3}}=\sqrt[3]{9}$

Solution: The variable appears in the base.

$& (2x-4)^{\frac{2}{3}}=\sqrt[3]{9} && \text{Apply} \ \boxed{a^{\frac{m}{n}}=\sqrt[n]{a^m}=\left(\sqrt[n]{a}\right)^m \ m,n \in N} \ \text{to the right side of the equation}.\\& (2x-4)^{\frac{2}{3}}=(9)^{\color{red}\frac{1}{3}} && \text{Write} \ 9 \ \text{as a power of} \ 3.\\& (2x-4)^{\frac{2}{3}}=({\color{red}3^2})^{\frac{1}{3}} && \text{Apply the law of exponents} \ \boxed{(a^m)^n=a^{mn}} \ \text{to the right side of the equation}.\\& (2x-4)^{\frac{2}{3}}=(3)^{\color{red}2 \times \frac{1}{3}} && \text{Simplify the exponents}.\\& (2x-4)^{\frac{2}{3}}=(3)^{\color{red}\frac{2}{3}} && \text{The exponents are equal so the bases are equal quantities}.\\& 2x-4=3 && \text{Solve the equation}.\\& 2x-4 {\color{red}+4}=3 {\color{red}+4}\\& 2x={\color{red}7}\\& \frac{2x}{{\color{red}2}}=\frac{7}{{\color{red}2}}\\& \frac{\cancel{2}x}{\cancel{2}}={\color{red}\frac{7}{2}}\\& \boxed{x=\frac{7}{2}}$

#### Concept Problem Revisited

$9^{x+1}=\sqrt{27}$

To begin, write each side of the equation with a common base. Both 9 and 27 can be written as a power of ‘3’. Therefore, $({\color{red}3^2})^{x+1}=\sqrt{{\color{red}3^3}}$ .

Apply $\boxed{(a^m)^n=a^{mn}}$ to the left side of the equation. $3^{{\color{red}2x+2}}=\sqrt{3^3}$

Express the right side of the equation in exponential form and apply $\boxed{(a^m)^n=a^{mn}}$ .

$& 3^{2x+2}=(3^3)^{{\color{red}\frac{1}{2}}}\\& 3^{2x+2}=3^{{\color{red}\frac{3}{2}}}$

Now that the bases are the same, then the exponents are equal quantities.

${\color{red}2x+2}={\color{red}\frac{3}{2}}$

Solve the equation.

${\color{red}2}(2x+2)={\color{red}2} \left(\frac{3}{2}\right)$ Multiply both sides of the equation by ‘2’. Simplify and solve.

$& {\color{red}4x+4}=\cancel{2} \left(\frac{{\color{red}3}}{\cancel{2}}\right)\\& 4x+4=3\\& 4x+4 {\color{red}-4}=3{\color{red}-4}\\& \frac{\cancel{4}x}{\cancel{4}}={\color{red}\frac{-1}{4}}\\& \boxed{x=-\frac{1}{4}}$

### Vocabulary

Exponential Equation
An exponential equation is an equation in which the variable appears in either the exponent or in the base. The equation is solved by applying the laws of exponents.

### Guided Practice

1. Use the laws of exponents to solve the following exponential equation: $27^{1-x}=\left(\frac{1}{9}\right)^{2-x}$

2. Use the laws of exponents to solve the following exponential equation: $(x-3)^{\frac{1}{2}}=(25)^{\frac{1}{4}}$

3. Use the laws of exponents to solve $\frac{(8^{x-4})(2^x)(4^{2x+3})}{32^x}=16$ .

1.

$& 27^{1-x}=\left(\frac{1}{9}\right)^{2-x} && \text{The variable appears in the exponent}.\\& ({\color{red}3^3})^{1-x}=({\color{red}3^{-2}})^{2-x} && \text{Write each side of the equation as a power of} \ 3.\\& (3^3)^{1-x}=(3^{-2})^{2-x} && \text{Apply the law of exponents} \ \boxed{(a^m)^n=a^{mn}}.\\& (3)^{{\color{red}3(1-x)}}=(3)^{{\color{red}-2(2-x)}} && \text{Simplify the exponents}.\\& 3^{{\color{red}3-3x}}=3^{{\color{red}-4+2x}} && \text{The bases are the same so the exponents are equal quantities}.\\& 3-3x=-4+2x && \text{Solve the equation}.\\& 3{\color{red}-3}-3x=-4{\color{red}-3}+2x\\& -3x={\color{red}-7}+2x\\& -3x{\color{red}-2x}=-7x+2x{\color{red}-2x}\\& {\color{red}-5x}=-7\\& \frac{-5x}{{\color{red}5}}=\frac{-7}{{\color{red}5}}\\& \frac{\cancel{-5}x}{\cancel{-5}}={\color{red}\frac{7}{5}}\\& \boxed{x=\frac{7}{5}}$

2.

$& (x-3)^{\frac{1}{2}}=(25)^{\frac{1}{4}} && \text{The variable appears in the base}.\\& (x-3)^{\frac{1}{2}}=({\color{red}5^2})^{\frac{1}{4}} && \text{Write} \ 25 \ \text{as a power of} \ 2.\\& (x-3)^{\frac{1}{2}}=({\color{red}5^2})^{\frac{1}{4}} && \text{Apply the law of exponents} \ \boxed{(a^m)^n=a^{mn}} \ \text{to the right side of the equation}.\\& (x-3)^{\frac{1}{2}}=(5)^{{\color{red}2 \times \frac{1}{4}}} && \text{Simplify the exponents}.\\& (x-3)^{\frac{1}{2}}=(5)^{{\color{red}\frac{2}{4}}}\\& (x-3)^{\frac{1}{2}}=(5)^{{\color{red}\frac{1}{2}}} && \text{The exponents are equal so the bases are equal quantities}.\\& x-3=5 && \text{Solve the equation}.\\& x-3 {\color{red}+3}=5 {\color{red}+3}\\& x={\color{red}8}\\& \boxed{x=8}$

3.

$& \frac{(8^{x-4})(2^x)(4^{2x+3})}{32^x}=16 && \text{The variable appears in the exponent}.\\& \frac{[({\color{red}2^3})^{x-4}](2^x)[({\color{red}2^2})^{2x+3}]}{({\color{red}2^5})^x}=2^{\color{red}4} && \text{Write all bases as a power of} \ 2. \ \text{Write} \ 16 \ \text{as a power of} \ 2.\\& \frac{[({\color{red}2^3})^{x-4}](2^x)[({\color{red}2^2})^{2x+3}]}{({\color{red}2^5})^x}=2^{\color{red}4} && \text{Apply the law of exponents} \ \boxed{(a^m)^n=a^{mn}}.\\ & \frac{[(2)^{{\color{red}3(x-4)}}](2^x)[(2)^{{\color{red}2(2x+3)}}]}{(2)^{{\color{red}5(x)}}}=2^{\color{red}4}\\& \frac{[(2)^{{\color{red}3x-12}}](2^x)[(2)^{{\color{red}4x+6}}]}{2^{\color{red}5x}}=2^{\color{red}4} && \text{Simplify the exponents}.\\& \frac{[2^{{\color{red}3x-12}}](2^x)[2^{{\color{red}4x+6}}]}{2^{{\color{red}5x}}}=2^{{\color{red}4}} && \text{Apply the law of exponents} \ \boxed{a^m \times a^n=a^{m+n}}.\\& \frac{[2^{{\color{red}3x+x+4x-12+6}}]}{2^{{\color{red}5x}}}=2^{\color{red}4} && \text{Simplify the exponents}.\\ & \frac{2^{{\color{red}8x-6}}}{2^{{\color{red}5x}}}=2^{{\color{red}4}} && \text{Apply the laws of exponents} \ \boxed{\frac{a^m}{a^n}=a^{m-n}}.\\& 2^{{\color{red}8x-6-5x}}=2^{\color{red}4} && \text{Simplify the exponents}.\\& 2^{{\color{red}3x-6}}=2^{\color{red}4} && \text{The bases are the same so the exponents are equal quantities}.\\& 3x-6=4 && \text{Solve the equation}.\\& 3x-6 {\color{red}+6}=4 {\color{red}+6}\\& 3x={\color{red}10}\\& \frac{3x}{{\color{red}3}}=\frac{10}{{\color{red}3}}\\& \frac{\cancel{3}x}{\cancel{3}}={\color{red}\frac{10}{3}}\\& \boxed{x=\frac{10}{3}}$

### Practice

Use the laws of exponents to solve the following exponential equations:

1. $2^{3x-1}=\sqrt[3]{16}$
2. $36^{x-2}=\left(\frac{1}{6}\right)^{2x+5}$
3. $6(x-4)^{\frac{1}{3}}=18$
4. $(3x-2)^{\frac{2}{5}}=4$
5. $36^{x+1}=\sqrt{6}$
6. $3^{5x-1}=\sqrt[3]{9}$
7. $9^{2x-1}=\left(\sqrt[4]{27}\right)^x$
8. $(3x-2)^{\frac{3}{2}}=8$
9. $(x+1)^{-\frac{5}{2}}=32$
10. $\left(\sqrt{3}\right)^{4x}=27^{x-3}$
11. $4^{3x-1}=\sqrt[3]{32}$
12. $(x+2)^{\frac{2}{3}}=(27)^{\frac{2}{9}}$
13. $(2^{x-3})(8^x)=32$
14. $(x-2)^{\frac{1}{2}}=9^{\frac{1}{4}}$
15. $8^{x+12}=\left(\frac{1}{16}\right)^{2x-7}$