Exponents and Rational Numbers - ACT Math

Card 0 of 90

Question

Which of the following is a value of that satisfies $\log_m256 =4$ ?

Answer

When you have a logarithm in the form

$y=\log_bx$ ,

it is equal to

.

Using the information given, we can rewrite the given equation in the second form to get

.

Now solving for we get the result.

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Question

Solve for :

$\log _{a}243 = 5$

Answer

When you have a logarithm in the form

$\log _{b}x = y$ ,

it is equal to

.

We can rewrite the given equation as

Solving for , we get

$\sqrt[5]{a^5} = \sqrt[5]{243}$

.

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Question

Solve for :

$\log_{4}a = 5$

Answer

When you have a logarithm in the form

$\log _{b}x = y$ ,

it is equal to

.

We can rewrite the given equation as

Solving for , we get

.

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Question

Sometimes, seeing rational numbers makes it easier to understand an equation.

Convert the following into a rational number or numbers:

$\frac{5^{\frac{1}{3}}\cdot 5^{\frac{4}{5}}}{5^{\frac{1}{15}}}$

Answer

The rule for converting exponents to rational numbers is: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$ .

Even with this, it is easier to work the problem as far as we can with exponents, then switch to rational expression when we run out of room:

$\frac{5^{\frac{1}{3}}\cdot 5^{\frac{4}{5}}}{5^{\frac{1}{15}}} = \frac{5^{\frac{17}{15}}}{5^\frac{1}{15}} = 5^{\frac{17}{15}-\frac{1}{15}} = 5^{\frac{16}{15}}$

At last, we convert, and obtain $5^{\frac{16}{15}} = \sqrt[15]{5^{16}}$ .

Thus,

$\frac{5^{\frac{1}{3}}\cdot 5^{\frac{4}{5}}}{5^{\frac{1}{15}}} = \sqrt[15]{5^{16}}$ .

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Question

Converting exponents to rational numbers often allows for faster simplification of those numbers.

Which of the following is incorrect? Convert exponents to rational numbers.

Answer

To identify which answer is incorrect we need to do each of the conversions.

First lets look at $\sqrt[4]{625} = 25^{\frac{1}{2}}$

$625^{\frac{1}{4}}=5$

$25^{\frac{1}{2}}=5$ .

Therefore this conversion is true.

Next lets look at $42^{\frac{0}{3}} = \sqrt[5]{1^{30}}$ . For this particular one we can recognize that anything raised to a zero power is just one therefore this conversion is true.

From here lets look at $16^{\frac{1}{2}} = \sqrt[3]{8}$

$16^{\frac{1}{2}} = \sqrt{16}$

$\sqrt{16} = 4$

$\sqrt[3]{8} = 2$

Thus

$\sqrt{16} eq \sqrt[3]{8}$ . Therefore this is an incorrect conversion and thus our answer.

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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 of the following is a value of that satisfies $\log_m256 =4$ ?

Answer

When you have a logarithm in the form

$y=\log_bx$ ,

it is equal to

.

Using the information given, we can rewrite the given equation in the second form to get

.

Now solving for we get the result.

Compare your answer with the correct one above

Question

Solve for :

$\log _{a}243 = 5$

Answer

When you have a logarithm in the form

$\log _{b}x = y$ ,

it is equal to

.

We can rewrite the given equation as

Solving for , we get

$\sqrt[5]{a^5} = \sqrt[5]{243}$

.

Compare your answer with the correct one above

Question

Solve for :

$\log_{4}a = 5$

Answer

When you have a logarithm in the form

$\log _{b}x = y$ ,

it is equal to

.

We can rewrite the given equation as

Solving for , we get

.

Compare your answer with the correct one above

Question

Sometimes, seeing rational numbers makes it easier to understand an equation.

Convert the following into a rational number or numbers:

$\frac{5^{\frac{1}{3}}\cdot 5^{\frac{4}{5}}}{5^{\frac{1}{15}}}$

Answer

The rule for converting exponents to rational numbers is: $a^{\frac{m}{n}} = \sqrt[n]{a^m}$ .

Even with this, it is easier to work the problem as far as we can with exponents, then switch to rational expression when we run out of room:

$\frac{5^{\frac{1}{3}}\cdot 5^{\frac{4}{5}}}{5^{\frac{1}{15}}} = \frac{5^{\frac{17}{15}}}{5^\frac{1}{15}} = 5^{\frac{17}{15}-\frac{1}{15}} = 5^{\frac{16}{15}}$

At last, we convert, and obtain $5^{\frac{16}{15}} = \sqrt[15]{5^{16}}$ .

Thus,

$\frac{5^{\frac{1}{3}}\cdot 5^{\frac{4}{5}}}{5^{\frac{1}{15}}} = \sqrt[15]{5^{16}}$ .

Compare your answer with the correct one above

Question

Converting exponents to rational numbers often allows for faster simplification of those numbers.

Which of the following is incorrect? Convert exponents to rational numbers.

Answer

To identify which answer is incorrect we need to do each of the conversions.

First lets look at $\sqrt[4]{625} = 25^{\frac{1}{2}}$

$625^{\frac{1}{4}}=5$

$25^{\frac{1}{2}}=5$ .

Therefore this conversion is true.

Next lets look at $42^{\frac{0}{3}} = \sqrt[5]{1^{30}}$ . For this particular one we can recognize that anything raised to a zero power is just one therefore this conversion is true.

From here lets look at $16^{\frac{1}{2}} = \sqrt[3]{8}$

$16^{\frac{1}{2}} = \sqrt{16}$

$\sqrt{16} = 4$

$\sqrt[3]{8} = 2$

Thus

$\sqrt{16} eq \sqrt[3]{8}$ . Therefore this is an incorrect conversion and thus our answer.

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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