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Algebra II : Solving and Graphing Exponential Equations

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Example Questions

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 #3871 : Algebra Ii

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

Possible Answers:

$-\frac{2}{7}$

$-\frac{1}{6}$

$\frac{2}{3}$

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 #3871 : Algebra Ii

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

Possible Answers:

$-\frac{1}{2}$

$\frac{1}{5}$

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 #101 : Solving And Graphing Exponential Equations

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

Possible Answers:

$\frac{2}{3}$

$\frac{1}{6}$

$\textup{There is no solution.}$

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:

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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 #1211 : Mathematical Relationships And Basic Graphs

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

Possible Answers:

$-\frac{5}{3}$

$\frac{1}{3}$

$-\frac{1}{4}$

$\frac{7}{3}$

$\frac{2}{9}$

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 #3872 : Algebra Ii

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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Example Question #101 : Solving Exponential Equations

Solve: $2^{5x+2} = (\frac{1}{2})^{10x+3}$

Possible Answers:

$\frac{1}{15}$

$\frac{1}{12}$

$-\frac{1}{6}$

$-\frac{1}{3}$

$-\frac{1}{9}$

Correct answer:

$-\frac{1}{3}$

Explanation:

Change the base of the right side.

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

Rewrite the base of the right side.

$2^{5x+2} = (2^{-1})^{10x+3}$

Set the exponents equal to each other now that both of the bases are alike.

5x+2 = -1(10x+3)

Simplify both sides.

5x+2 = -10x-3

Add 10x on both sides.

5x+2 +10x= -10x-3+10x

15x+2= -3

Subtract two on both sides.

15x+2-2= -3-2

15x=-5

Divide by 15 on both sides.

$\frac{15x}{15}=\frac{-5}{15}$

Reduce both fractions.

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

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Example Question #102 : Solving Exponential Equations

Solve the equation: $8^{10x} = 8^{-9x-19}$

Possible Answers:

$\textup{There is no solution.}$

$\frac{19}{8}$

Correct answer:

Explanation:

The bases on both sides of the equation are equal. This means we can set the powers equal to each other.

10x= -9x-19

Add on both sides.

10x+9x= -9x-19+9x

19x = -19

Divide by 19 on both sides.

$\frac{19x}{19} = \frac{-19}{19}$

The answer is:

Report an Error

Example Question #101 : Solving Exponential Equations

Solve the equation: $4^{3x+2} = (\frac{1}{4})^{-9x+3}$

Possible Answers:

$-\frac{5}{12}$

$\textup{No solution.}$

$\frac{5}{12}$

$\frac{5}{3}$

$\frac{5}{6}$

Correct answer:

$\frac{5}{6}$

Explanation:

Change the base on the right side to match the left side.

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

$4^{3x+2} = (4)^{-1(-9x+3)}$

Set the exponents equal to each other since both bases are equal.

3x+2 = -1(-9x+3)

Simplify both sides.

3x+2 =9x-3

Subtract both sides by and add both sides by three.

3x+2 -(3x)+[3]=9x-3-(3x)+[3]

5=6x

Divide both sides by six.

$\frac{5}{6}=\frac{6x}{6}$

The answer is: $\frac{5}{6}$

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Algebra II : Solving and Graphing Exponential Equations

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #94 : Solving Exponential Equations

Example Question #3871 : Algebra Ii

Example Question #3871 : Algebra Ii

Example Question #101 : Solving And Graphing Exponential Equations

Example Question #98 : Solving Exponential Equations

Example Question #1211 : Mathematical Relationships And Basic Graphs

Example Question #3872 : Algebra Ii

Example Question #101 : Solving Exponential Equations

Example Question #102 : Solving Exponential Equations

Example Question #101 : Solving Exponential Equations

All Algebra II Resources

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