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Algebra II : Solving Exponential Equations

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Example Question #91 : Solving Exponential Equations

Solve the equation: $\frac{1}{e^{-3x+6}} = e^{6x}$

Possible Answers:

$-\frac{3}{4}$

$\textup{No solution.}$

$-\frac{1}{2}$

Correct answer:

Explanation:

We can rewrite the left side of the equation as follows:

$\frac{1}{e^{-3x+6}} =( \frac{1}{e})^{-3x+6} = (e^{-1})^{-3x+6}$

Rewrite the equation.

$(e^{-1})^{-3x+6} = e^{6x}$

Since both sides share the same bases, we can set the powers equal to each other.

(-1)(-3x+6) =6x

Solve for x.

3x-6 = 6x

Subtract on both sides.

3x-6 -3x= 6x-3x

-6 =3x

Divide by three on both sides.

$\frac{-6 }{3}=\frac{3x}{3}$

The answer is:

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Example Question #92 : Solving Exponential Equations

Solve: $3^x = (\frac{1}{81})^{9x-1}$

Possible Answers:

$-\frac{9}{23}$

$-\frac{4}{23}$

$\frac{14}{37}$

$\frac{4}{37}$

$-\frac{4}{35}$

Correct answer:

$\frac{4}{37}$

Explanation:

Rewrite the right side using three as the base.

$(\frac{1}{81})^{9x-1} = (3^{-4})^{9x-1}$

The equation becomes:

$3^x = (3)^{-4(9x-1)}$

Since both sides share similar bases, we can set the powers equal to each other.

x=-36x+4

Add 36x on both sides.

x+36x=-36x+4+36x

37x = 4

Divide both sides by 37.

$\frac{37x }{37}= \frac{4}{37}$

The answer is: $\frac{4}{37}$

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Example Question #93 : Solving Exponential Equations

Solve: $(\frac{1}{2})^{5x+2} =( \frac{1}{8})^{3x}$

Possible Answers:

$\frac{1}{2}$

$\textup{There is no solution.}$

$\frac{3}{2}$

$\frac{3}{4}$

Correct answer:

$\frac{1}{2}$

Explanation:

We can rewrite the right side by changing the base.

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

Rewrite the right side using the new base.

$(\frac{1}{2})^{5x+2} =[(\frac{1}{2})^3]^{3x}$

Now that the bases are similar, we can set the exponents equal to each other.

5x+2 = 9x

Subtract on both sides.

5x+2 -5x= 9x-5x

2=4x

Divide by four on both sides.

$\frac{2}{4}=\frac{4x}{4}$

Reduce the fractions.

The answer is: $\frac{1}{2}$

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Example Question #94 : Solving Exponential Equations

Solve the equation: $3(\frac{1}{3})^{3x+1}= (\frac{1}{9})^{10}$

Possible Answers:

$\textup{There is no solution.}$

$\frac{40}{3}$

$\frac{19}{3}$

$\frac{20}{3}$

Correct answer:

$\frac{20}{3}$

Explanation:

We will need to convert the bases on both sides to base three.

$\frac{1}{3} = 3^{-1}$

$\frac{1}{9}=(\frac{1}{3})^2$

Rewrite the equation.

$3[(3)^{-1}]^{3x+1}= (3^{-2})^{10}$

Simplify both sides.

$3(3)^{-1(3x+1)}= (3^{-2})^{10}$

On the left side, notice that a lone three is the coefficient of the base. We will need to write the left side such that:

$3(3)^{-1(3x+1)} = 3^1\cdot (3)^{-1(3x+1)} = 3^{1+(-1)(3x+1)}$

Recall that if the powers of the same base are multiplied, the powers can be added.

Simplify the exponent.

$3^{1+(-1)(3x+1)} = 3^{1-3x-1} = 3^{-3x}$

The equation becomes: $3^{-3x} = 3^{-20}$

Set the exponential terms equal to each other now that the bases are equal.

-3x=-20

Divide by negative three on both sides.

$\frac{-3x}{-3}=\frac{-20}{-3}$

The answer is: $\frac{20}{3}$

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Example Question #95 : Solving Exponential Equations

Solve: $2^{3x+3} = 8^{-5x}$

Possible Answers:

$\frac{2}{3}$

$-\frac{2}{7}$

$-\frac{1}{6}$

Correct answer:

$-\frac{1}{6}$

Explanation:

Change the base of the second term to base two.

8=2^3

$2^{3x+3} =(2^3)^{-5x}$

Simplify the right side by multiplying the exponents.

$2^{3x+3} =(2)^{-15x}$

Now that the bases are common, the exponents can be set equal to each other.

3x+3 = -15x

Subtract from both sides.

3x+3-3x = -15x-3x

3=-18x

Divide by negative 18 on both sides.

$\frac{3}{-18}=\frac{-18x}{-18}$

The answer is: $-\frac{1}{6}$

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Example Question #96 : Solving Exponential Equations

Solve: $4^6 = 2^{3x-3}$

Possible Answers:

$\frac{1}{5}$

$-\frac{1}{2}$

Correct answer:

Explanation:

To evaluate this equation, we will need to change the base of the left side.

4=2^2

Rewrite the equation.

$(2^2)^6 = 2^{3x-3}$

$2^{12}= 2^{3x-3}$

Now that both bases are similar, we can set the exponents equal to each other.

12=3x-3

Add 3 on both sides.

12+3=3x-3+3

15=3x

Divide by three on both sides.

The answer is:

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Example Question #97 : Solving Exponential Equations

Solve the equation: $2(\frac{1}{2})^{7x-1}= (\frac{1}{4})^{3x+2}$

Possible Answers:

$\frac{2}{3}$

$\textup{There is no solution.}$

$\frac{1}{6}$

Correct answer:

Explanation:

Rewrite both sides of the equation so that we have same bases.

$2(\frac{1}{2})^{7x-1}= (\frac{1}{4})^{3x+2}$

2=2^1

$\frac{1}{2} = 2^{-1}$

$\frac{1}{4} = 2^{-2}$

$2^1\cdot (2^{-1})^{7x-1}= (2^{-2})^{3x+2}$

Simplify the exponents.

$2^1\cdot (2^{-7x+1})= 2^{-6x-4}$

Add the exponents on the left side.

$2^{-7x+2}= 2^{-6x-4}$

Now that both sides have same bases, we can set the exponential terms equal.

-7x+2 = -6x-4

Add on both sides.

-7x+2+7x = -6x-4+7x

2 = x-4

Add four on both sides.

2 +4= x-4+4

The answer is:

Report an Error

Example Question #98 : Solving Exponential Equations

Solve: $3^4 = (\frac{1}{27})^{9x-3}$

Possible Answers:

$-\frac{2}{3}$

$\frac{1}{9}$

$\frac{5}{27}$

$-\frac{13}{27}$

$\frac{5}{3}$

Correct answer:

$\frac{5}{27}$

Explanation:

In order to solve this, we will need to rewrite the fractional base to base three.

$\frac{1}{27} = 3^{-3}$

Rewrite the equation using this base.

$3^4 = (3^{-3})^{9x-3}$

Now that both bases are common, we can set up an equation where the powers are equal.

4=-3(9x-3)

Use distribution to simplify the right side.

4=-27x+9

Subtract nine from both sides.

4-9=-27x+9-9

-5 = -27x

Divide by negative 27 on both sides.

$\frac{-5 }{-27}= \frac{-27x}{-27}$

The answer is: $\frac{5}{27}$

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Example Question #99 : Solving Exponential Equations

Evaluate: $(\frac{2}{3})^{3x-8} =( \frac{8}{27})^{-x-2}$

Possible Answers:

$\frac{7}{3}$

$\frac{2}{9}$

$-\frac{1}{4}$

$\frac{1}{3}$

$-\frac{5}{3}$

Correct answer:

$\frac{1}{3}$

Explanation:

To solve this exponential equation, we will need to change the base of the second term. Notice that both the numerator and denominator are the values of the left fraction cubed.

$\frac{8}{27} = (\frac{2}{3})^3$

Rewrite the equation.

$(\frac{2}{3})^{3x-8} =( \frac{2}{3})^{3(-x-2)}$

With common bases, we can set the exponents equal to each other.

3x-8 = 3(-x-2)

Distribute the right side.

3x-8 = -3x-6

Add on both sides.

3x-8 +(3x)= -3x-6+(3x)

6x-8 = -6

Add 8 on both sides.

6x-8 +8= -6+8

6x=2

Divide by six on both sides.

$\frac{6x}{6}=\frac{2}{6}$

The answer is: $\frac{1}{3}$

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Example Question #100 : Solving Exponential Equations

Solve: $(\frac{1}{9}) ^{81} = 3^{2x-10}$

Possible Answers:

$\textup{No solution.}$

Correct answer:

Explanation:

In order to solve this equation, we will need to change the base of the left side.

Rewrite one-ninth as base three.

$\frac{1}{9} = 3^{-2}$

Rewrite the equation.

$(3^{-2}) ^{81} = 3^{2x-10}$

Now that both bases are common, we can set both powers equal.

(-2)(81) = 2x-10

Simplify the left side and solve for x.

-162 = 2x-10

Add 10 on both sides, and then divide by two.

-162 +10= 2x-10+10

$\frac{-152}{2}=\frac{2x}{2}$

The answer is:

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Algebra II : Solving Exponential Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #91 : Solving Exponential Equations

Example Question #92 : Solving Exponential Equations

Example Question #93 : Solving Exponential Equations

Example Question #94 : Solving Exponential Equations

Example Question #95 : Solving Exponential Equations

Example Question #96 : Solving Exponential Equations

Example Question #97 : Solving Exponential Equations

Example Question #98 : Solving Exponential Equations

Example Question #99 : Solving Exponential Equations

Example Question #100 : Solving Exponential Equations

All Algebra II Resources

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