### About Solving Exponential Equations with Logarithms:

When we work with exponential equations, we have an easy case, where we can find our solution without the use of logarithms. This happens when we have the same base on each side of the equation. The harder case occurs when we can't write each side with the same base. For this scenario, we turn to logarithms and our power property to obtain a solution.

Test Objectives

- Demonstrate an understanding of the properties of logarithms
- Demonstrate the ability to solve an exponential equation using logarithms

#1:

Instructions: solve each equation.

$$a)\hspace{.2em}18^{6x}+ 6=60$$

$$b)\hspace{.2em}{-}10 \cdot 15^{4x + 10}- 4=25$$

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#2:

Instructions: solve each equation.

$$a)\hspace{.2em}2 \cdot 20^{-6x - 9}- 7=19$$

$$b)\hspace{.2em}4^{x}+ 4^{2 - x}=10$$ Hint: Use the rules of exponents, then make a substitution.

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#3:

Instructions: solve each equation.

$$a)\hspace{.2em}9^{x - 1}=3^{x + 4}$$

$$b)\hspace{.2em}3^x + 9^x=90$$ Hint: Use the rules of exponents, then make a substitution.

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#4:

Instructions: solve each equation.

$$a)\hspace{.2em}e^{5x - 1}\cdot e^{2x}=7e$$

$$b)\hspace{.2em}(4x + 3)^{x^2 - 16}=1$$ Hint: Suppose you have a^{b} = 1. What are the possible values for a and b using only real numbers?

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#5:

Instructions: solve each word problem.

a) If money is invested in a CD that is compounded continuously at 1.9% annual interest, how long will it take for the investment to double in value?

Continuous Compound Interest Formula: $$A=Pe^{rt}$$

b) If money is invested in a CD that is compounded continuously at 3.5% annual interest, how long will it take for the investment to triple in value?

Continuous Compound Interest Formula: $$A=Pe^{rt}$$

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Written Solutions:

#1:

Solutions:

$$a)\hspace{.2em}x=\frac{\text{log}(54)}{6\text{log}(18)}\approx 0.23$$

$$b)\hspace{.2em}\text{No Solution}$$

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#2:

Solutions:

$$a)\hspace{.2em}x=-\frac{3}{2}- \frac{\text{log}(13)}{6\text{log}(20)}\approx -1.6427$$

$$b)\hspace{.2em}x=\frac{1}{2}, \frac{3}{2}$$

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#3:

Solutions:

$$a)\hspace{.2em}x=6$$

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

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#4:

Solutions:

$$a)\hspace{.2em}x=\frac{\text{ln}(7) + 2}{7}\approx 0.5637$$

$$b)\hspace{.2em}x=\pm 4, -\frac{1}{2}$$

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#5:

Solutions:

a) About 36.48 Years

b) About 31.39 Years