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# Solving Exponential Equations

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Practice Solving Exponential Equations
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Solving Exponential Equations

"I'm thinking of a number," you tell your best friend. "The number I'm thinking of satisfies the equation $4^{x + 1} = 256$ . What number are you thinking of?

### Guidance

Until now, we have only solved pretty basic exponential equations, like #1 in the Review Queue above. We know that $x=5$ , because $2^5=32$ . Ones like #4 are a little more challenging, but if we put everything into a power of 2, we can set the exponents equal to each other and solve.

$8^x &= 128 \\2^{3x} &= 2^7 \\3x &= 7 \\x &= \frac{7}{3}$

So, $8^{\frac{7}{3}} = 128$ .

But, what happens when the power is not easily found? We must use logarithms, followed by the Power Property to solve for the exponent.

#### Example A

Solve $6^x=49$ . Round your answer to the nearest three decimal places.

Solution: To solve this exponential equation, let’s take the logarithm of both sides. The easiest logs to use are either $\ln$ (the natural log), or log (log, base 10). We will use the natural log.

$6^x &= 49 \\\ln 6^x &= \ln 49 \\x \ln 6 &= \ln 49 \\x &= \frac{\ln 49}{\ln 6} \approx 2.172$

#### Example B

Solve $10^{x-3}=100^{3x+11}$ .

Solution: Change 100 into a power of 10.

$10^{x-3} &= 10^{2(3x+11)} \\x-3 &= 6x+22 \\-25 &= 5x \\-5 &= x$

#### Example C

Solve $8^{2x-3}-4=5$ .

Solution: Add 4 to both sides and then take the log of both sides.

$8^{2x-3}-4 &= 5 \\8^{2x-3} &= 9 \\\log 8^{2x-3} &= \log 9 \\(2x-3)\log 8 &= \log 9 \\2x-3 &= \frac{\log 9}{\log 8} \\2x &= 3 + \frac{\log 9}{\log 8} \\x &= \frac{3}{2}+\frac{\log 9}{2 \log 8} \approx 2.56$

Notice that we did not find the numeric value of $\log9$ or $\log8$ until the very end. This will ensure that we have the most accurate answer.

Intro Problem Revisit We can rewrite the equation $4^{x + 1} = 256$ as $2^{2(x+1)} = 2^8$ and solve for x .

$2^{2(x+1)} = 2^8\\2^{2x +2} = 2^8\\2x + 2 = 8\\x = 3$

Therefore, you're thinking of the number 3.

### Guided Practice

Solve the following exponential equations.

1. $4^{x-8}=16$

2. $2(7)^{3x+1} =48$

3. $\frac{2}{3} \cdot 5^{x+2}+9=21$

1. Change 16 to $4^2$ and set the exponents equal to each other.

$4^{x-8} &= 16 \\4^{x-8} &= 4^2 \\x-8 &= 2 \\x &=10$

2. Divide both sides by 2 and then take the log of both sides.

$2(7)^{3x+1} &= 48 \\7^{3x+1} &= 24 \\\ln 7^{3x+1} &= \ln 24 \\(3x+1)\ln 7 &= \ln 24 \\3x+1 &= \frac{\ln 24}{\ln 7} \\3x &= -1 + \frac{\ln 24}{\ln 7} \\x &= -\frac{1}{3} + \frac{\ln 24}{3 \ln 7} \approx 0.211$

3. Subtract 9 from both sides and multiply both sides by $\frac{3}{2}$ . Then, take the log of both sides.

$\frac{2}{3} \cdot 5^{x+2}+9 &= 21 \\\frac{2}{3} \cdot 5^{x+2} &= 12 \\5^{x+2} &= 18 \\(x+2)\log 5 &= \log 18 \\x &= \frac{\log 18}{\log 5}-2 \approx -0.204$

### Practice

Use logarithms and a calculator to solve the following equations for $x$ . Round answers to three decimal places.

1. $5^x = 65$
2. $7^x = 75$
3. $2^x = 90$
4. $3^{x-2} = 43$
5. $6^{x+1}+3=13$
6. $6(11^{3x-2})=216$
7. $8+13^{2x-5}=35$
8. $\frac{1}{2} \cdot 7^{x-3}-5=14$

Solve the following exponential equations without a calculator.

1. $4^x=8$
2. $9^{x-2} = 27$
3. $5^{2x+1}=125$
4. $9^3=3^{4x-6}$
5. $7(2^{x-3})=56$
6. $16^x \cdot 4^{x+1}=32^{x+1}$
7. $3^{3x+5}=3 \cdot 9^{x+3}$