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Algebra II : Simplifying Exponents

Study concepts, example questions & explanations for Algebra II

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Example Questions

Example Question #141 : Simplifying Exponents

Simplify: $\frac{x^7}{x^8}$

Possible Answers:

$x^{15}$

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

$\frac{1}{x}$

$-\frac{1}{x}$

Correct answer:

$\frac{1}{x}$

Explanation:

When dividing exponents, we subtract the exponents and keep the base the same.

$\frac{x^7}{x^8}=x^{7-8}=x^{-1}$ We know with negative exponents, it's expressed as one over the positive exponent.

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

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Example Question #3602 : Algebra Ii

Simplify: $x^5*(\frac{5}{x^4})^2*\frac{x^8}{10}$

Possible Answers:

$\frac{5x^4}{2}$

$\frac{x^9}{2}$

x^5

$\frac{5x^9}{2}$

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

Correct answer:

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

Explanation:

Let's apply the exponents to the parentheses first and then simplify.

$x^5*(\frac{5}{x^4})^2*\frac{x^8}{10}=x^5*\frac{25}{x^8}*\frac{x^8}{10}$ The x^8 cancels and the numbers can be reduced by .

We finally get: $\frac{5x^5}{2}$ .

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Example Question #142 : Simplifying Exponents

Simplify and express as exponents: $\frac{x^4}{y^7}*x^{-8}$

Possible Answers:

(xy)^7

$x^{-4}y^7$

(xy)^4

x^4y^7

$x^{-4}y^{-7}$

Correct answer:

$x^{-4}y^{-7}$

Explanation:

Let's rewrite this as just exponents. Remember we can breakup $\frac{x^4}{y^7}=x^4*\frac{1}{y^7}$ . $\frac{x^4}{y^7}*x^{-8}=\frac{1}{y^7}*x^4*x^{-8}=y^{-7}*x^{-4}$

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Example Question #143 : Simplifying Exponents

Simplify: $\frac{5^{2x}}{5^{2x+4}}$

Possible Answers:

$-\frac{1}{5^4}$

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

-5^4

5^4

Correct answer:

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

Explanation:

When dividing exponents, we subtract the exponents and keep the base the same.

$\frac{5^{2x}}{5^{2x+4}}=5^{2x-(2x+4)}=5^{2x-2x-4}=5^{-4}=\frac{1}{5^4}$ We know with negative exponents, it's expressed as one over the positive exponent.

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Example Question #144 : Simplifying Exponents

Simplify: $\frac{7^{8x+4}}{7^{7x-2}}$

Possible Answers:

$\frac{1}{7^{x+2}}$

$-\frac{1}{7^{x+6}}$

$7^{x+2}$

$\frac{1}{7^{x+6}}$

$7^{x+6}$

Correct answer:

$7^{x+6}$

Explanation:

When dividing exponents, we subtract the exponents and keep the base the same.

$\frac{7^{8x+4}}{7^{7x-2}}=7^{8x+4-(7x-2)}=7^{8x+4-7x+2}=7^{x+6}$

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Example Question #146 : Multiplying And Dividing Exponents

Simplify: $\frac{a+b}{a^2+2ab+b^2}$

Possible Answers:

a-b

$\frac{1}{2ab}$

$\frac{1}{a+b}$

a+b

$\frac{1}{a+ab+b}$

Correct answer:

$\frac{1}{a+b}$

Explanation:

Although it seems like we can't simplify anything, we do know that a^2+2ab+b^2=(a+b)^2 . Therefore we have:

$\frac{a+b}{a^2+2ab+b^2}=\frac{a+b}{(a+b)^2}$ . Now we can divide the exponents to get $(a+b)^{-1}=\frac{1}{a+b}$

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Example Question #147 : Multiplying And Dividing Exponents

Simplify: $(\frac{x}{y})^{7n}*(\frac{x}{y^{-1}})^{2n}*\frac{y^{5n}}{x^{5n+2}}$

Possible Answers:

$x^{4n+2}$

$\frac{x^{4n}}{y^{2n}}$

$\frac{x^{4n-2}}{y^n}$

$\frac{x^n}{y^n}$

$x^{4n-2}$

Correct answer:

$x^{4n-2}$

Explanation:

Let's apply the exponential operation before we simplify.

$(\frac{x}{y})^{7n}*(\frac{x}{y^{-1}})^{2n}*\frac{y^{5n}}{x^{5n+2}}=\frac{x^{7n}}{y^{7n}}*x^{2n}y^{2n}*\frac{y^{5n}}{x^{5n+2}}$ In the numerator, the $y^{2n}, y^{5n}$ become $y^{7n}$ and cancels with the denominator in the left fraction.

We now have: $x^{7n}*x^{2n}*\frac{1}{x^{5n+2}}$ . By combining the top and applying the division rule of exponents, we get:

$x^{7n}*x^{2n}*\frac{1}{x^{5n+2}}=\frac{x^{9n}}{x^{5n+2}}=x^{9n-(5n+2)}=x^{9n-5n-2}=x^{4n-2}$

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Example Question #145 : Simplifying Exponents

Simplify: $\frac{2^{12}}{4^6}$

Possible Answers:

$-\frac{1}{4}$

$\frac{1}{2}$

$\frac{1}{4}$

Correct answer:

Explanation:

Although the exponents have different bases, we know that 4=2^2 . Therefore we can rewrite as $\frac{2^{12}}{4^6}=\frac{2^{12}}{(2^2)^6}=\frac{2^{12}}{2^{12}}=1$

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Example Question #149 : Multiplying And Dividing Exponents

Simplify: $\frac{2^{24}8^8}{4^{16}}$

Possible Answers:

2^8

$2^{12}$

$\frac{1}{2^8}$

$2^{16}$

$\frac{1}{2^{12}}$

Correct answer:

$2^{16}$

Explanation:

Although the bases are not the same, we know that 8=2^3, 4=2^2 . We will base our answers in base of since this is present in all the choices. Therefore: $\frac{2^{24}8^8}{4^{16}}=\frac{2^{24}(2^3)^8}{(2^2)^{16}}=\frac{2^{24}2^{24}}{2^{32}}$ Now we add the exponents and then subtract them.

$\frac{2^{24}2^{24}}{2^{32}}=\frac{2^{48}}{2^{32}}=2^{16}$

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Example Question #150 : Multiplying And Dividing Exponents

Divide: $\frac{1}{5}\div 25^{-10}$

Possible Answers:

$\frac{5}{25^{15}}$

$5^{21}$

$25^{15}$

$\frac{5}{25^{21}}$

$5^{19}$

Correct answer:

$5^{19}$

Explanation:

In order to divide, we will first need to change the bases of both terms.

Both terms can be rewritten as base five.

$\frac{1}{5}\div 25^{-10} = 5^{-1} \div 5^{2(-10)} = \frac{5^{-1}}{5^{-20}}$

Now that bases are common, we can subtract the powers and the base will be left unchanged.

$5^{-1-(-20)} = 5^{19}$

The answer is: $5^{19}$

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Algebra II : Simplifying Exponents

Study concepts, example questions & explanations for Algebra II

All Algebra II Resources

Example Questions

Example Question #141 : Simplifying Exponents

Example Question #3602 : Algebra Ii

Example Question #142 : Simplifying Exponents

Example Question #143 : Simplifying Exponents

Example Question #144 : Simplifying Exponents

Example Question #146 : Multiplying And Dividing Exponents

Example Question #147 : Multiplying And Dividing Exponents

Example Question #145 : Simplifying Exponents

Example Question #149 : Multiplying And Dividing Exponents

Example Question #150 : Multiplying And Dividing Exponents

All Algebra II Resources

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