Negative Exponents

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Algebra II › Negative Exponents

Questions 1 - 10

Simplify $(4x^{2}y^{-6})(2x^{-7}y^{^3})$ .

$\frac{8}{x^{5}y^{3}}$

$\frac{8}{x^{14}y^{18}}$

$8x^{5}y^{3}$

$\frac{1}{8x^{5}y^{3}}$

$\frac{1}{8x^{14}y^{18}}$

Explanation

First multiply the like terms, remembering that when multiplying terms that have exponents, you add the exponents.

$(4x^{2}y^{-6})(2x^{-7}y^{3})=(4\cdot 2)(x^{2}\cdot x^{-7})(y^{-6}\cdot y^{3})$

$=8x^{2-7}y^{-6+3}$

$=8x^{-5}y^{-3}$

Negative exponents indicate that the term should be in the denominator, so the final answer is:

$\frac{8}{x^{5}y^{3}}$

Solve for :

$(x+5)^{-3} = -1$

Explanation

Raise both sides of the equation to the inverse power of to cancel the exponent on the left hand side of the equation.

$\rightarrow ((x+5)^{-3})^{-\frac{1}{3}} = (-1)^{-\frac{1}{3}}$

$\rightarrow x+5 = -1$

Subtract from both sides:

$\rightarrow (x+5) - 5 = (-1)-5$

$\rightarrow x = -6$

Evaluate: $3^{-4}$

$\frac{1}{81}$

$-\frac{1}{81}$

Explanation

When dealing with exponents, we convert as such: $x^{-a}=\frac{1}{x^a}$ . Therefore, $3^{-4}=\frac{1}{3^4}=\frac{1}{81}$ .

Evaluate: $7^{-3}$

$\frac{1}{7^3}$

$-\frac{1}{7^3}$

Explanation

When dealing with negative exponents, we write $x^{-a}=\frac{1}{x^a}$ . Therefore $7^{-3}=\frac{1}{7^3}$ .

Simplify:

$10^{-5}$

$\frac{1}{10^5}$

$-\frac{1}{10^5}$

Explanation

When an exponent is negative, we express as such:

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

is the positive exponent, and is the base.

$10^{-5}=\frac{1}{10^5}= \frac{1}{100000}$ .

Evaluate: $5^{-8}$

$\frac{1}{5^8}$

$-\frac{1}{5^8}$

$\frac{1}{5^{-8}}$

$-5^{-8}$

Explanation

When dealing with negative exponents, we convert to fractions as such: $x^{-a}=\frac{1}{x^a}$ which is the positive exponent raising base .

$5^{-8}=\frac{1}{5^8}$

Evaluate $2^{-1}$

$\frac{1}{2}$

$-\frac{1}{2}$

Explanation

When dealing with exponents, always turn it into this form:

$\frac{1}{x^n}$ represents the base of the exponent, and is the power in a positive value.

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

Evaluate: $2^{-3x}=8$

$-\frac{8}{3}$

$-\frac{2}{3}$

$-\frac{1}{4}$

$\textup{The answer is not given.}$

Explanation

In order to determine the value of x, we will need to convert the base of the right side similar to the left.

Eight is similar to two cubed. Rewrite the equation.

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

Now that our bases are the same, we can set the exponents equal to each other.

Divide by negative three on both sides.

$\frac{-3x}{-3}=\frac{3}{-3}$

The answer is:

Evaluate: $4^{-4}$

$\frac{1}{256}$

$-\frac{1}{256}$

$-\frac{1}{16}$

$\frac{1}{16}$

Explanation

When dealing with exponents, we convert as such: $x^{-a}=\frac{1}{x^a}$ . Therefore $4^{-4}=\frac{1}{4^4}=\frac{1}{256}$

Simplify: $(ab^2c)^{-3}$

$\frac{1}{a^3b^6c^3}$

$\frac{1}{a^3bc^3}$

$\frac{b}{a^3c^3}$

$\frac{b^2}{a^3c^3}$

$\frac{b^6}{a^3c^3}$

Explanation

The terms as a quantity can be rewritten by the following property:

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

Rewrite the expression.

$(ab^2c)^{-3} = \frac{1}{(ab^2c)^3}$

Expand the terms by distributing the third power with every term in the denominator.

The answer is: $\frac{1}{a^3b^6c^3}$

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