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Algebra II : Exponents

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Example Questions

Example Question #81 : Solving Exponential Equations

Solve for .

$2^{x-9}=(\frac{1}{2})^{x+1}$

Possible Answers:

Correct answer:

Explanation:

When dealing with exponential equations, we want to make sure the bases are the same. This way we can set-up an equation with the exponents. Since the bases are now different, we need to convert so we have the same base. We do know that

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

$2^{x-9}=(2^{-1})^{x+1}$ With the same base and also applying power rule of exponents, we now can write

x-9=-x-1 Add x, 9 on both sides.

2x=8 Divide on both sides.

x=4

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Example Question #81 : Solving Exponential Equations

Solve for .

$4^{2x+3}=(\frac{1}{4})^{-x+6}$

Possible Answers:

Correct answer:

Explanation:

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

$4^{2x+3}=(4^{-1})^{-x+6}$ With the same base and by also applying the power rule of exponents, we can now write

2x+3=x-6 Subtract x, 3 on both sides.

x=-9

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Example Question #82 : Solving Exponential Equations

Solve: $(\frac{1}{3})^{8x} = (\frac{1}{81})^{3-x}$

Possible Answers:

$\frac{2}{3}$

$-\frac{1}{4}$

Correct answer:

Explanation:

In order to solve this equation, we will need to change the bases so that both bases are equal. We can change the bases to base three.

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

$\frac{1}{81} = 3^{-4}$

Rewrite the terms in the equation.

$(3^{-1})^{8x} = (3^{-4})^{3-x}$

The powers can be set equal to each other now that we have similar bases.

(-1)(8x) = (-4)(3-x)

Solve for x.

-8x = -12+4x

Subtract on both sides.

-8x-4x = -12+4x-4x

-12x=-12

Divide by negative 12 on both sides.

$\frac{-12x}{-12}=\frac{-12}{-12}$

The answer is:

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Example Question #83 : Solving Exponential Equations

Evaluate: $(\frac{1}{16})^{9x} = 16^{3+x}$

Possible Answers:

$-\frac{5}{4}$

$\textup{No solution.}$

$-\frac{3}{10}$

$-\frac{3}{8}$

$-\frac{5}{8}$

Correct answer:

$-\frac{3}{10}$

Explanation:

Evaluate by changing the base of the fraction.

$\frac{1}{16} = 16^{-1}$

Rewrite the equation.

$(16)^{-1(9x)} = 16^{3+x}$

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

-1(9x)=3+x

Solve for x.

-9x = 3+x

Subtract x from both sides.

-9x -x= 3+x-x

-10x = 3

Divide by negative ten on both sides.

$\frac{-10x }{-10}= \frac{3}{-10}$

The answer is: $-\frac{3}{10}$

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Example Question #731 : Exponents

Solve the equation: $7^{x-1} = 49^{\frac{1}{x}}$

Possible Answers:

$\textup{No solution.}$

-2,-1

-1,2

Correct answer:

-1,2

Explanation:

Convert the right side to a base of 7.

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

With similar bases, the powers can be set equal to each other.

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

Multiply both sides by .

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

x^2-x =2

Subtract two from both sides.

x^2-x -2=2-2

x^2-x -2=0

Factorize the left term to determine the roots of the quadratic.

(x+1)(x-2)=0

Set each binomial equal to zero and solve for x.

x+1 =0, x=-1

x-2=0, x=2

Ensure that both solutions will satisfy the original equation. Since there is no zero denominator resulting from substitution, both answers will satisfy.

The answers are: -1,2

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Example Question #3865 : Algebra Ii

Solve: $(\frac{2}{3})^{-5x} =(\frac{4}{9})^{2+x}$

Possible Answers:

$-\frac{3}{8}$

$-\frac{4}{7}$

$\frac{2}{9}$

$\textup{No solution.}$

$\frac{3}{2}$

Correct answer:

$-\frac{4}{7}$

Explanation:

Convert the right fraction to a base of two thirds.

$\frac{4}{9} =(\frac{2}{3})^2$

Rewrite the right side.

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

With similar bases, set both powers equal to each other.

-5x = 2(2+x)

Simplify the right side by distribution.

-5x= 4+2x

Subtract from both sides.

-5x-2x= 4+2x-2x

-7x =4

Divide both sides by negative seven.

$\frac{-7x }{-7}=\frac{4}{-7}$

The answer is: $-\frac{4}{7}$

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Example Question #92 : Solving And Graphing Exponential Equations

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

Possible Answers:

$-\frac{1}{2}$

$\textup{No solution.}$

$-\frac{3}{4}$

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 #93 : Solving And Graphing Exponential Equations

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

Possible Answers:

$\frac{14}{37}$

$\frac{4}{37}$

$-\frac{4}{35}$

$-\frac{4}{23}$

$-\frac{9}{23}$

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 #91 : Solving Exponential Equations

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

Possible Answers:

$\frac{1}{2}$

$\frac{3}{2}$

$\frac{3}{4}$

$\textup{There is no solution.}$

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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Algebra II : Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #81 : Solving Exponential Equations

Example Question #81 : Solving Exponential Equations

Example Question #82 : Solving Exponential Equations

Example Question #83 : Solving Exponential Equations

Example Question #731 : Exponents

Example Question #3865 : Algebra Ii

Example Question #92 : Solving And Graphing Exponential Equations

Example Question #93 : Solving And Graphing Exponential Equations

Example Question #91 : Solving Exponential Equations

Example Question #94 : Solving Exponential Equations

All Algebra II Resources

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