How to find an exponent from a rational number - ACT Math

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Question

Which real number satisfies $2^{n}\cdot 4=8^{2}$ ?

Answer

Simplying the equation, we get $2^{n}\cdot 4=64$ .

This further simplifies to $2^{n}=16$ .

$n=4$ satisfies this equation. You could also use $\log _{2}16$ to determine that $2^{4}=16$ .

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Question

Simplify by expressing each term in exponential form.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{(8^{\frac{1}{3}})(x^{\frac{7}{3}})}{(81^{\frac{1}{4}})(x^{\frac{2}{4}})} = \frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}}$

The middle step in the coversion is important, as there is a big difference between $8^{\frac{1}{3}}$ and $8^{\frac{7}{3}}$ , and likewise for the denominator.

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}} = \frac{2x^{(\frac{7}{3}-\frac{1}{2})}}{3} = \frac{2x^{\frac{11}{6}}}{3}$

Thus,

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{2x^{\frac{11}{6}}}{3}$

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Question

Find the value of if $\frac{64}{27}=\left ( \frac{4}{3} \right )^{x}$

Answer

When a fraction is raised to an exponent, both the numerator and denominator are raised to that exponent. Therefore, the equation can be rewritten as $\frac{64}{27}=\frac{4^{x}}{3^{x}}$ . From here we can proceed one of two ways. We can either solve x for $64=4^{x}$ or $27=3^{x}$ . Let's solve the first equation. We simply multiply 4 by itself until we reach a value of 64. $4^{1}=4$ , $4^{2}=4\cdot4=16$ , $4^{3}=4\cdot4\cdot4=64$ , and so on. Since $4^{3}=64$ , we know that x = 3.

We can repeat this process for the second equation to get $3^{3}=3\cdot3\cdot3= 27$ , confirming our previous answer. However, since the ACT is a timed test, it is best to only solve one of the equations and move on. Then, if you have time left once all of the questions have been answered, you can come back and double check your answer by solving the other equation.

Compare your answer with the correct one above

Question

If $\small f(x)=2\times x^{3}$ and $\small g(x)=x+2$ , what is $\small f(g(3))$ ?

Answer

Start from the inside. $\small g(3)=3+2=5$ . Then, $\small f(5)=2\times 5^{3}=2\times 125=250$ .

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Question

Often, solving a root equation is as simple as switching to exponential form.

Simplify into exponential form:

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}} = \frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}}$

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}} = 125^{\frac{5}{15}-\frac{3}{15}} = 125^{\frac{2}{15}}$

Thus, our answer is $125^{\frac{2}{15}}$ . (Remember, the problem asked for exponential form!)

Compare your answer with the correct one above

Question

Which real number satisfies $2^{n}\cdot 4=8^{2}$ ?

Answer

Simplying the equation, we get $2^{n}\cdot 4=64$ .

This further simplifies to $2^{n}=16$ .

$n=4$ satisfies this equation. You could also use $\log _{2}16$ to determine that $2^{4}=16$ .

Compare your answer with the correct one above

Question

Simplify by expressing each term in exponential form.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{(8^{\frac{1}{3}})(x^{\frac{7}{3}})}{(81^{\frac{1}{4}})(x^{\frac{2}{4}})} = \frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}}$

The middle step in the coversion is important, as there is a big difference between $8^{\frac{1}{3}}$ and $8^{\frac{7}{3}}$ , and likewise for the denominator.

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}} = \frac{2x^{(\frac{7}{3}-\frac{1}{2})}}{3} = \frac{2x^{\frac{11}{6}}}{3}$

Thus,

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{2x^{\frac{11}{6}}}{3}$

Compare your answer with the correct one above

Question

Find the value of if $\frac{64}{27}=\left ( \frac{4}{3} \right )^{x}$

Answer

When a fraction is raised to an exponent, both the numerator and denominator are raised to that exponent. Therefore, the equation can be rewritten as $\frac{64}{27}=\frac{4^{x}}{3^{x}}$ . From here we can proceed one of two ways. We can either solve x for $64=4^{x}$ or $27=3^{x}$ . Let's solve the first equation. We simply multiply 4 by itself until we reach a value of 64. $4^{1}=4$ , $4^{2}=4\cdot4=16$ , $4^{3}=4\cdot4\cdot4=64$ , and so on. Since $4^{3}=64$ , we know that x = 3.

We can repeat this process for the second equation to get $3^{3}=3\cdot3\cdot3= 27$ , confirming our previous answer. However, since the ACT is a timed test, it is best to only solve one of the equations and move on. Then, if you have time left once all of the questions have been answered, you can come back and double check your answer by solving the other equation.

Compare your answer with the correct one above

Question

If $\small f(x)=2\times x^{3}$ and $\small g(x)=x+2$ , what is $\small f(g(3))$ ?

Answer

Start from the inside. $\small g(3)=3+2=5$ . Then, $\small f(5)=2\times 5^{3}=2\times 125=250$ .

Compare your answer with the correct one above

Question

Often, solving a root equation is as simple as switching to exponential form.

Simplify into exponential form:

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}} = \frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}}$

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}} = 125^{\frac{5}{15}-\frac{3}{15}} = 125^{\frac{2}{15}}$

Thus, our answer is $125^{\frac{2}{15}}$ . (Remember, the problem asked for exponential form!)

Compare your answer with the correct one above

Question

Which real number satisfies $2^{n}\cdot 4=8^{2}$ ?

Answer

Simplying the equation, we get $2^{n}\cdot 4=64$ .

This further simplifies to $2^{n}=16$ .

$n=4$ satisfies this equation. You could also use $\log _{2}16$ to determine that $2^{4}=16$ .

Compare your answer with the correct one above

Question

Simplify by expressing each term in exponential form.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{(8^{\frac{1}{3}})(x^{\frac{7}{3}})}{(81^{\frac{1}{4}})(x^{\frac{2}{4}})} = \frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}}$

The middle step in the coversion is important, as there is a big difference between $8^{\frac{1}{3}}$ and $8^{\frac{7}{3}}$ , and likewise for the denominator.

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}} = \frac{2x^{(\frac{7}{3}-\frac{1}{2})}}{3} = \frac{2x^{\frac{11}{6}}}{3}$

Thus,

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{2x^{\frac{11}{6}}}{3}$

Compare your answer with the correct one above

Question

Find the value of if $\frac{64}{27}=\left ( \frac{4}{3} \right )^{x}$

Answer

When a fraction is raised to an exponent, both the numerator and denominator are raised to that exponent. Therefore, the equation can be rewritten as $\frac{64}{27}=\frac{4^{x}}{3^{x}}$ . From here we can proceed one of two ways. We can either solve x for $64=4^{x}$ or $27=3^{x}$ . Let's solve the first equation. We simply multiply 4 by itself until we reach a value of 64. $4^{1}=4$ , $4^{2}=4\cdot4=16$ , $4^{3}=4\cdot4\cdot4=64$ , and so on. Since $4^{3}=64$ , we know that x = 3.

We can repeat this process for the second equation to get $3^{3}=3\cdot3\cdot3= 27$ , confirming our previous answer. However, since the ACT is a timed test, it is best to only solve one of the equations and move on. Then, if you have time left once all of the questions have been answered, you can come back and double check your answer by solving the other equation.

Compare your answer with the correct one above

Question

If $\small f(x)=2\times x^{3}$ and $\small g(x)=x+2$ , what is $\small f(g(3))$ ?

Answer

Start from the inside. $\small g(3)=3+2=5$ . Then, $\small f(5)=2\times 5^{3}=2\times 125=250$ .

Compare your answer with the correct one above

Question

Often, solving a root equation is as simple as switching to exponential form.

Simplify into exponential form:

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}} = \frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}}$

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}} = 125^{\frac{5}{15}-\frac{3}{15}} = 125^{\frac{2}{15}}$

Thus, our answer is $125^{\frac{2}{15}}$ . (Remember, the problem asked for exponential form!)

Compare your answer with the correct one above

Question

Which real number satisfies $2^{n}\cdot 4=8^{2}$ ?

Answer

Simplying the equation, we get $2^{n}\cdot 4=64$ .

This further simplifies to $2^{n}=16$ .

$n=4$ satisfies this equation. You could also use $\log _{2}16$ to determine that $2^{4}=16$ .

Compare your answer with the correct one above

Question

Simplify by expressing each term in exponential form.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{(8^{\frac{1}{3}})(x^{\frac{7}{3}})}{(81^{\frac{1}{4}})(x^{\frac{2}{4}})} = \frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}}$

The middle step in the coversion is important, as there is a big difference between $8^{\frac{1}{3}}$ and $8^{\frac{7}{3}}$ , and likewise for the denominator.

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{2x^{\frac{7}{3}}}{3x^{\frac{1}{2}}} = \frac{2x^{(\frac{7}{3}-\frac{1}{2})}}{3} = \frac{2x^{\frac{11}{6}}}{3}$

Thus,

$\frac{\sqrt[3]{8x^{7}}}{\sqrt[4]{81x^2}} = \frac{2x^{\frac{11}{6}}}{3}$

Compare your answer with the correct one above

Question

Find the value of if $\frac{64}{27}=\left ( \frac{4}{3} \right )^{x}$

Answer

When a fraction is raised to an exponent, both the numerator and denominator are raised to that exponent. Therefore, the equation can be rewritten as $\frac{64}{27}=\frac{4^{x}}{3^{x}}$ . From here we can proceed one of two ways. We can either solve x for $64=4^{x}$ or $27=3^{x}$ . Let's solve the first equation. We simply multiply 4 by itself until we reach a value of 64. $4^{1}=4$ , $4^{2}=4\cdot4=16$ , $4^{3}=4\cdot4\cdot4=64$ , and so on. Since $4^{3}=64$ , we know that x = 3.

We can repeat this process for the second equation to get $3^{3}=3\cdot3\cdot3= 27$ , confirming our previous answer. However, since the ACT is a timed test, it is best to only solve one of the equations and move on. Then, if you have time left once all of the questions have been answered, you can come back and double check your answer by solving the other equation.

Compare your answer with the correct one above

Question

If $\small f(x)=2\times x^{3}$ and $\small g(x)=x+2$ , what is $\small f(g(3))$ ?

Answer

Start from the inside. $\small g(3)=3+2=5$ . Then, $\small f(5)=2\times 5^{3}=2\times 125=250$ .

Compare your answer with the correct one above

Question

Often, solving a root equation is as simple as switching to exponential form.

Simplify into exponential form:

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}}$

Answer

The rule for exponential ratios is $a^{\frac{m}{n}} = \sqrt[n]{a^{m}}$ .

Using this, we can convert the numerator and denominator quickly.

$\frac{\sqrt[3]{125}}{\sqrt[5]{125}} = \frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}}$

Next, we can further simplify by remembering that

$\frac{a^{m}}{a^{n}} = a^{m-n}$ .

Find the least common denominator and simplify:

$\frac{125^{\frac{1}{3}}}{125^{\frac{1}{5}}} = 125^{\frac{5}{15}-\frac{3}{15}} = 125^{\frac{2}{15}}$

Thus, our answer is $125^{\frac{2}{15}}$ . (Remember, the problem asked for exponential form!)

Compare your answer with the correct one above

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