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College Algebra : Review and Other Topics

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Example Questions

Example Question #2 : Real Exponents

Simplify the following expression.

$(2x^{7.2})(x^{-6.2})$

Possible Answers:

2x^2

$2x^{13.2}$

$x^{1.2}$

Correct answer:

Explanation:

When multiplying exponential, the exponents always add. While the 2 in the front of the first exponential might throw you off, you may disregard it initially.

$(2)(x^{7.2}x^{-6.2})$

Which simplifies to

$(2)x^{7.2-6.2}=(2)x^1$

Our final answer is

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Example Question #4 : Real Exponents

Simplify the following expression:

$\frac{(x^{2})^{5}x^{-8}}{x^2}$

Possible Answers:

x^3

x^2

Correct answer:

Explanation:

First, we need to simplify the numerator. First term, (x^2)^5 can be simplified to $x^{10}$ . Plugging this back into the numerator, we get

$x^{10}x^{-8}$ , which simplifies to x^2 . Plugging this back into the original equation gives us

$\frac{x^2}{x^2}$ , which is simply .

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Example Question #4 : Real Exponents

Solve for :

$a^{2}\cdot a^{3}= 243$

Possible Answers:

a=6

a=2

a=5

a=3

a=4

Correct answer:

a=3

Explanation:

When like bases with exponents are multiplied, the value of the product's exponent is the sum of both original exponents as shown here:

$a^{n}\cdot a^{m}=a^{n+m}$

We can use this common rule to solve for in the practice problem:

$\\a^{2}\cdot a^{3}=243\\a^{2}\cdot a^{3}=a^{2+3}\\a^{5}=243$

$a=\sqrt[5]{243}\\a=3$

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Example Question #4 : Real Exponents

Solve for :

$\frac{a^{7}}{a^3}=256$

Possible Answers:

a=6

a=3

a=7

a=4

a=5

Correct answer:

a=4

Explanation:

The product of dividing like bases with exponents is the difference of the numerator and denominator exponents. This is a common rule when working with rational exponents:

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

We can use this common rule to solve for :

$\\\frac{a^{7}}{a^{3}}=256\\\\a^{7-3}=256\\a^{4}=256\\a=\sqrt[4]{256}\\a=4$

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Example Question #4 : Real Exponents

Solve for :

$3^{x-2}\cdot9^{2x}=27$

Possible Answers:

x=1

x=0

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

x=-1

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

Correct answer:

x=1

Explanation:

To solve for , we need all values to have like bases:

$\\3^{x-2}\cdot9^{2x}=27\\3^{x-2}\cdot3^{2\cdot2x}=27\\3^{x-2}\cdot3^{4x}=3^{3}$

Now that all values have like bases, we can solve for :

$\\x-2+4x=3\\5x-2=3\\5x=5\\x=1$

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Example Question #6 : Real Exponents

Solve for :

$27^{\frac{x-3}{3}}\cdot3^{\frac{3x}{4}}=81$

Possible Answers:

x=3

x=-4

x=4

x=0

x=-3

Correct answer:

x=4

Explanation:

To solve for , we want all values in the equation to have like bases:

$\\27^{\frac{x-3}{3}}\cdot3^{\frac{3x}{4}}=81 \\3^{3\cdot\frac{x-3}{3}}\cdot3^{\frac{3x}{4}}=3^{4} \\3^{x-3}\cdot3^{\frac{3x}{4}}=3^{4}$

Now we can solve for :

$\\x-3+\frac{3x}{4}=4\\\frac{7x}{4}=7\\7x=28\\x=4$

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Example Question #5 : Real Exponents

Solve for :

$5^{\frac{x}{4}}\cdot5^{\frac{x}{8}}=125$

Possible Answers:

x=4

x=16

x=12

x=24

x=8

Correct answer:

x=8

Explanation:

To solve for , we want all the values in the equation to have like bases:

$\\5^{\frac{x}{4}}\cdot5^{\frac{x}{8}}=125 \\5^{\frac{x}{4}}\cdot5^{\frac{x}{8}}=5^3$

Now we can solve for :

$\\\frac{x}{4}+\frac{x}{8}=3 \\8\cdot(\frac{x}{4}+\frac{x}{8})=3\cdot8 \\2x+x=24\\3x=24\\x=8$

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Example Question #6 : Real Exponents

Simplify the following expression.

$\frac{x^{\sqrt{2}}}{x^2}*\frac{x^{\sqrt{2}}}{x^2}$

Possible Answers:

x^2

$x^{-2}$

$x^{\sqrt{2}}$

Correct answer:

$x^{-2}$

Explanation:

The original expression can be rewritten as

$(\frac{x^{\sqrt{2}}}{x^2})^2$ . Whenever you can a fraction raised to a power, that power gets distributed out to the numerator and denominator. In mathematical terms, the new expression is

$\frac{x^{2}}{x^4}$ , which simply becomes $\frac{1}{x^2}$ , or $x^{-2}$

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Example Question #1 : Radicals

Subtract and simplify:
$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Possible Answers:

$\small \frac{\sqrt{60}-\sqrt{150}}{10}$

$\small \frac{2\sqrt{15}-5\sqrt6}{10}$

$\small 3(\sqrt2-\sqrt5)$

$\small \frac{\sqrt{6}-\sqrt{15}}{\sqrt{10}}$

Correct answer:

$\small \frac{2\sqrt{15}-5\sqrt6}{10}$

Explanation:

$\small \frac{\sqrt3}{\sqrt5}-\frac{\sqrt3}{\sqrt2}$

Find the lease common denominator: $\small \sqrt{10}$

$\small \small \frac{\sqrt{3}}{\sqrt5}=\frac{\sqrt6}{\sqrt{10}}$

$\small \small \frac{\sqrt{3}}{\sqrt2}=\frac{\sqrt{15}}{\sqrt{10}}$

$\small \small \frac{\sqrt6}{\sqrt10}-\frac{\sqrt{15}}{\sqrt{10}}=\frac{\sqrt6-\sqrt{15}}{\sqrt{10}}$

A radical cannot be in the denominator:

$\small \small (\frac{\sqrt6-\sqrt{15}}{\sqrt{10}})(\frac{\sqrt{10}}{\sqrt{10}})=\frac{\sqrt{60}-\sqrt{150}}{10}$

$\small \sqrt{60}=2\sqrt{15}$

$\small \sqrt{150}=5\sqrt6$

$\small \frac{\sqrt{60}-\sqrt{150}}{10}=\frac{2\sqrt{15}-5\sqrt6}{10}$

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Example Question #1 : Radicals

Find the value of $\sqrt{80} + \sqrt{20}$ .

Possible Answers:

$6\sqrt{10}$

$6\sqrt5$

$4\sqrt5$

$\sqrt5$

Correct answer:

$6\sqrt5$

Explanation:

To solve this equation, we have to factor our radicals. We do this by finding numbers that multiply to give us the number within the radical.

$\sqrt{80} = \sqrt{20}\sqrt{4}$

$\sqrt{20} = \sqrt{5}\sqrt{4}$

Add them together:

$\sqrt5\sqrt4\sqrt4 + \sqrt5\sqrt4$

4 is a perfect square, so we can find the root:

$(2)(2)\sqrt5 + 2\sqrt5$

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

Since both have the same radical, we can combine them:

$4\sqrt5 + 2\sqrt5 = 6\sqrt5$

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College Algebra : Review and Other Topics

Study concepts, example questions & explanations for College Algebra

All College Algebra Resources

Example Questions

Example Question #2 : Real Exponents

Example Question #4 : Real Exponents

Example Question #4 : Real Exponents

Example Question #4 : Real Exponents

Example Question #4 : Real Exponents

Example Question #6 : Real Exponents

Example Question #5 : Real Exponents

Example Question #6 : Real Exponents

Example Question #1 : Radicals

Example Question #1 : Radicals

All College Algebra Resources

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