Solving Exponential Equations - High School Math

Card 0 of 19

Question

Solve the equation for .

$\small 9^x=3^6$

Answer

Begin by recognizing that both sides of the equation have a root term of .

$\small 9^x=3^6$

Using the power rule, we can set the exponents equal to each other.

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

$\small 2x=6$

$\small x=3$

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Question

Solve the equation for .

$\small 3^{2x}=81$

Answer

Begin by recognizing that both sides of the equation have the same root term, .

$\small 3^{2x}=81$

$\small 3^{2x}=9^2$

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

We can use the power rule to combine exponents.

$3^{2x}=3^4$

Set the exponents equal to each other.

$\small x=2$

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Question

Which value for satisfies the equation $4^x = \frac{192}{x}$ ?

Answer

is the only choice from those given that satisfies the equation. Substition of for gives:

$4^3 = \frac{192}{3} \rightarrow 64 = 64$

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Question

Solve for :

Answer

To solve for in the equation

Factor out of the expression on the left of the equation:

$\rightarrow 2(x^2 - 16) =0$

Use the "difference of squares" technique to factor the parenthetical term on the left side of the equation.

$\rightarrow 2(x+4)(x-4)=0$

Any variable that causes any one of the parenthetical terms to become will be a valid solution for the equation. becomes when is , and becomes when is , so the solutions are and .

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Question

Solve for :

Answer

Pull an out of the left side of the equation.

$\rightarrow x(x^2 - 9) =0$

Use the difference of squares technique to factor the expression in parentheses.

$\rightarrow x(x+3)(x-3) =0$

Any number that causes one of the terms , , or to equal is a solution to the equation. These are , , and , respectively.

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Question

Solve for (nearest hundredth):

$8^{y} = 100$

Answer

Take the common logarithm of both sides and solve for :

$8^{y} = 100$

$\log 8^{y} = \log 100$

$y \log 8 =2$

$y =\frac{2}{\log 8} \approx \frac{2}{0.9031} \approx 2.21$

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Question

Solve for (nearest hundredth):

$4 ^{2x - 7} = \frac{1}{256}$

Answer

$\frac{1}{256} = \frac{1}{4 ^{4}} = 4 ^{-4}$ , so $4 ^{2x - 7} = \frac{1}{256}$ can be rewritten as

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

$\log_{4} 4 ^{2x - 7} =\log_{4} 4 ^{-4}$

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Question

Solve for (nearest hundredth):

$80 ^{t} = 39$

Answer

One method: Take the natural logarithm of both sides and solve for :

$80 ^{t} = 39$

$\ln 80 ^{t} = \ln 39$

$t \ln 80 = \ln 39$

$t =\frac{ \ln 39}{ \ln 80} \approx \frac{ 3.66}{ 4.38} \approx 0.84$

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Question

Solve for :

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

Answer

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

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

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

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

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

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

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

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Question

Solve for :

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

Answer

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

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

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

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

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

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

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Question

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

What are the x-intercepts of the equation?

Answer

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

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

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

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

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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Question

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

What are the y-intercepts of the equation?

Answer

To find the y-intercepts, set 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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Question

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

What are the x-intercepts of the equation?

Answer

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

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

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

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

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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Question

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

What are the y-intercepts of the equation?

Answer

To find the y-intercepts, set 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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Question

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

What are the y-intercepts of this equation?

Answer

To find the y-intercept, set 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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Question

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

What are the x-intercepts of this equation?

Answer

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

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Question

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

What are the x-intercepts of the equation?

Answer

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

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

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Question

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

What are the y-intercepts of this equation?

Answer

To find the y-intercept, set and solve.

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

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

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

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Question

The population of a certain bacteria increases exponentially according to the following equation:

$P(t)=2000e^{^{2t}}$

where P represents the total population and t represents time in minutes.

How many minutes does it take for the bacteria's population to reach 48,000?

Answer

The question gives us P (48,000) and asks us to find t (time). We can substitute for P and start to solve for t:

$48,000 = 2000e^{2t}$

$24 = e^{2t}$

Now we have to isolate t by taking the natural log of both sides:

$\ln 24 = \ln e^{2t}$

$\ln 24 = (2t)\ln e$

And since $\ln e = 1$ , t can easily be isolated:

$\ln 24 = 2t$

$\frac{\ln24}{2} = t$

Note: $\frac{\ln 24}{2}$ does not equal $\ln 12$ . You have to perform the log operation first before dividing.

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