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High School Math : Solving and Graphing Exponential Equations

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Example Question #1 : Solving Exponential Equations

Solve for :

$3 ^{4x + 7} = 9 ^{2x +3}$

Possible Answers:

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

x=0

x = 2

x = 1

The equation has no solution.

Correct answer:

The equation has no solution.

Explanation:

Since $9 = 3^{2}$ , we can rewrite this equation by subsituting and applying the power rule:

$3 ^{4x + 7} = 9 ^{2x +3}$

$3 ^{4x + 7} = \left (3 ^{2} \right ) ^{2x +3}$

$3 ^{4x + 7} =3 ^{2 \left ( 2x +3 \right )}$

$3 ^{4x + 7} =3 ^{4x +6}$

$\log_{3} 3 ^{4x + 7} = \log_{3} 3 ^{4x +6}$

4x + 7 = 4x + 6

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

7 = 6

This statement is identically false, which means that the original equation is identically false. There is no solution.

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Example Question #1 : Solving Exponential Equations

Solve for :

$\left (\frac{1}{4} \right) ^{x - 7} = 2^{2x +5}$

Possible Answers:

x = 4.5

The equation has no solution

x = -9.5

x = 2.25

x = -4.75

Correct answer:

x = 2.25

Explanation:

$\frac{1}{4} = \frac{1}{2^{2}} =2^{-2}$ , so we can rewrite the equation as follows:

$\left (\frac{1}{4} \right) ^{x - 7} = 2^{2x +5}$

$\left (2^{-2}) ^{x - 7} = 2^{2x +5}$

$2^{-2 \left ( x - 7 \right )} = 2^{2x +5}$

$2^{-2x +14 \right )} = 2^{2x +5}$

$\log_2 2^{-2x +14 \right )} =\log_2 2^{2x +5}$

-2x + 14 = 2x + 5

-2x + 14 +2x -5= 2x + 5+2x -5

9= 4x

x = 2.25

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Example Question #8 : Solving Exponential Equations

$y=\frac{3x^2-5}{2x^2+7}$

What are the y-intercepts of the equation?

Possible Answers:

y=0

This equation does not have a y-intercept.

$y=\frac{5}{}7$

$y=-\frac{5}{7}$

$y=\frac{6}{35}$

Correct answer:

$y=-\frac{5}{7}$

Explanation:

To find the y-intercepts, set x=0 and solve.

$y=\frac{3x^2-5}{2x^2+7}$

$y=\frac{3(0)^2-5}{2(0)^2+7}$

$y=\frac{-5}{7}$

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Example Question #9 : Solving Exponential Equations

$y=\frac{4x^2-2}{2x^3-27}$

What are the y-intercepts of the equation?

Possible Answers:

$y=\frac{2}{27}$

There are no y-intercepts for this equation.

$y=\frac{3}{2}$

y=0

$y=\frac{27}{2}$

Correct answer:

$y=\frac{2}{27}$

Explanation:

To find the y-intercepts, set x=0 and solve.

$y=\frac{4x^2-2}{2x^3-27}$

$y=\frac{4(0)^2-2}{2(0)^3-27}$

$y=\frac{-2}{-27}$

$y=\frac{2}{27}$

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Example Question #1 : Solving Exponential Equations

$y=\frac{x-3}{x^2-12}$

What are the x-intercepts of this equation?

Possible Answers:

x=3

$x=\frac{1}{4}$

x=4

x=3, -3

$x=2\sqrt{3}$

Correct answer:

x=3

Explanation:

To find the x-intercepts, set the numerator equal to zero.

0=x-3

3=x

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Example Question #131 : Algebra Ii

$y=\frac{4x^2-2}{2x^3-27}$

What are the x-intercepts of the equation?

Possible Answers:

$x=\sqrt{2}, -\sqrt{2}$

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

$x=\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$

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

$x=\sqrt{2}$

Correct answer:

$x=\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}$

Explanation:

To find the x-intercepts, set the numerator equal to zero and solve.

0=4x^2-2

2=4x^2

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

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

$\sqrt{\frac{1}{2}}=\sqrt{x^2}$

We can simplify from here:

$\frac{\sqrt1}{\sqrt2}}=\sqrt{x^2}$

$\frac{1}{\sqrt2}}=\sqrt{x^2}$

Now we need to rationalize. Because we have a square root on the bottom, we need to get rid of it. Since $2=\sqrt{2}^2$ , we can multiply $\frac{1}{\sqrt2}*\frac{\sqrt{2}}{\sqrt{2}}$ to get rid of the radical in the denominator.

$\frac{1}{\sqrt2}*\frac{\sqrt{2}}{\sqrt{2}}=x$

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

Since we took a square root, remember that our answer can be either positive or negative, as a positive squared is positive and a negative squared is also positive.

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Example Question #132 : Algebra Ii

$y=\frac{x-3}{x^2-12}$

What are the y-intercepts of this equation?

Possible Answers:

$y=2\sqrt{3}$

y=0

y=4

There are no y-intercepts.

$y=\frac{1}{4}$

Correct answer:

$y=\frac{1}{4}$

Explanation:

To find the y-intercept, set x=0 and solve.

$y=\frac{x-3}{x^2-12}$

$y=\frac{(0)-3}{(0)^2-12}$

$y=\frac{-3}{-12}$

$y=\frac{1}{4}$

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Example Question #133 : Algebra Ii

$y=\frac{x^2-64}{x+2}$

What are the y-intercepts of this equation?

Possible Answers:

There are no y-intercepts for the equation.

y=32

y=0

y=-32

y=-2

Correct answer:

y=-32

Explanation:

To find the y-intercept, set x=0 and solve.

$y=\frac{x^2-64}{x+2}$

$y=\frac{(0)^2-64}{0+2}$

$y=\frac{-64}{2}$

y=-32

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Example Question #134 : Algebra Ii

$y=\frac{3x^2-5}{2x^2+7}$

What are the x-intercepts of the equation?

Possible Answers:

$x=\sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}$

x=0

x=1

$x=\sqrt{\frac{5}{3}}$

There are no horizontal asymptotes.

Correct answer:

$x=\sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}$

Explanation:

To find the x-intercepts, we set the numerator equal to zero and solve.

0=3x^2-5

5=3x^2

$\frac{5}{3}=x^2$

$\sqrt{\frac{5}{3}}=\sqrt{x^2}$

$\sqrt{\frac{5}{3}}=x$

However, the square root of a number can be both positive and negative.

Therefore the roots will be $x=\sqrt{\frac{5}{3}}, -\sqrt{\frac{5}{3}}.$

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Example Question #135 : Algebra Ii

$y=\frac{x^2-64}{x+2}$

What are the x-intercepts of the equation?

Possible Answers:

There are no x-intercepts.

There are no real x-intercepts.

x=8, -8

x=-2

x=0

Correct answer:

x=8, -8

Explanation:

To find the x-intercepts, set the numerator equal to zero.

0=x^2-64

64=x^2

$\sqrt{64}=\sqrt{x^2}$

8,-8=x

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High School Math : Solving and Graphing Exponential Equations

Study concepts, example questions & explanations for High School Math

All High School Math Resources

Example Questions

Example Question #1 : Solving Exponential Equations

Example Question #1 : Solving Exponential Equations

Example Question #8 : Solving Exponential Equations

Example Question #9 : Solving Exponential Equations

Example Question #1 : Solving Exponential Equations

Example Question #131 : Algebra Ii

Example Question #132 : Algebra Ii

Example Question #133 : Algebra Ii

Example Question #134 : Algebra Ii

Example Question #135 : Algebra Ii

All High School Math Resources

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