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Evaluating Exponential Expressions

Evaluate numbers raised to positive, negative, and fractional powers

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Evaluating Exponential Expressions

What if you had an exponential expression requiring multiple operations, like \begin{align*}2\left(\frac{1}{4}\right)^2 - \left(\frac{1}{4}\right)^3\end{align*}? How could you simplify it? After completing this Concept, you'll be able to use the order of operations to evaluate exponential expressions like this one.

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CK-12 Foundation: 0807S Evaluating Exponential Expressions

Guidance

When evaluating expressions we must keep in mind the order of operations. You must remember PEMDAS:

  1. Evaluate inside the Parentheses.
  2. Evaluate Exponents.
  3. Perform Multiplication and Division operations from left to right.
  4. Perform Addition and Subtraction operations from left to right.

Example A

Evaluate the following expressions.

a) \begin{align*}5^0\end{align*}

b) \begin{align*}\left(\frac{2}{3}\right)^3\end{align*}

c) \begin{align*}16^{\frac{1}{2}}\end{align*}

d) \begin{align*}8^{-\frac{1}{3}}\end{align*}

Solution

a) \begin{align*}5^0=1\end{align*} A number raised to the power 0 is always 1.

b) \begin{align*}\left(\frac{2}{3}\right)^3=\frac{2^3}{3^3}=\frac{8}{27}\end{align*}

c) \begin{align*}16^{\frac{1}{2}}=\sqrt{16}=4\end{align*} Remember that an exponent of \begin{align*}\frac{1}{2}\end{align*} means taking the square root.

d) \begin{align*}8^{-\frac{1}{3}}=\frac{1}{8^{\frac{1}{3}}}=\frac{1}{\sqrt[3]{8}}=\frac{1}{2}\end{align*} Remember that an exponent of \begin{align*}\frac{1}{3}\end{align*} means taking the cube root.

Example B

Evaluate the following expressions.

a) \begin{align*}3 \cdot 5^2-10 \cdot 5+1\end{align*}

b) \begin{align*}\frac{2 \cdot 4^2-3 \cdot 5^2}{3^2-2^2}\end{align*}

c) \begin{align*}\left(\frac{3^3}{2^2}\right)^{-2} \cdot \frac{3}{4}\end{align*}

Solution

a) Evaluate the exponent: \begin{align*}3 \cdot 5^2 - 10 \cdot 5+1=3 \cdot 25-10 \cdot 5+1\end{align*}

Perform multiplications from left to right: \begin{align*}3 \cdot 25-10 \cdot 5+1=75-50+1\end{align*}

Perform additions and subtractions from left to right: \begin{align*}75-50+1=26\end{align*}

b) Treat the expressions in the numerator and denominator of the fraction like they are in parentheses: \begin{align*}\frac{(2 \cdot 4^2-3 \cdot 5^2)}{(3^2-2^2)}=\frac{(2 \cdot 16-3 \cdot 25)}{(9-4)}=\frac{(32-75)}{5}=\frac{-43}{5}\end{align*}

c) \begin{align*}\left(\frac{3^3}{2^2}\right)^{-2} \cdot \frac{3}{4}=\left(\frac{2^2}{3^3}\right)^2 \cdot \frac{3}{4}=\frac{2^4}{3^6} \cdot \frac{3}{4}=\frac{2^4}{3^6} \cdot \frac{3}{2^2}=\frac{2^2}{3^5}=\frac{4}{243}\end{align*}

Example C

Evaluate the following expressions for \begin{align*}x = 2, y = - 1, z = 3\end{align*}.

a) \begin{align*}2x^2-3y^3+4z\end{align*}

b) \begin{align*}(x^2-y^2)^2\end{align*}

c) \begin{align*}\left(\frac{3x^2y^5}{4z}\right)^{-2}\end{align*}

Solution

a) \begin{align*}2x^2-3y^3+4z&=2 \cdot 2^2-3 \cdot (-1)^3+4 \cdot 3\\ &=2 \cdot 4-3 \cdot (-1)+4 \cdot 3=8+3+12\\ &=23\end{align*}

b) \begin{align*}(x^2-y^2)^2=(2^2 - (-1)^2)^2=(4-1)^2=3^2=9\end{align*}

c)

\begin{align*}\left( \frac{3x^2y^5}{4z}\right)^{-2}&=\left( \frac{3 \cdot 2^2 \cdot (-1)^5}{4 \cdot 3}\right)^{-2}\\ &=\left(\frac{3 \cdot 4 \cdot (-1)}{12}\right)^{-2}\\ &=\left(\frac{-12}{12}\right)^{-2}\\ &=\left(\frac{-1}{1}\right)^{-2}\\ &=\left(\frac{1}{-1}\right)^2\\ &=(-1)^2\\ &=1\end{align*}

Watch this video for help with the Examples above.

CK-12 Foundation: Evaluating Exponential Expressions

Vocabulary

  • When evaluating expressions we must keep in mind the order of operations. You must remember PEMDAS:
  1. Evaluate inside the Parentheses.
  2. Evaluate Exponents.
  3. Perform Multiplication and Division operations from left to right.
  4. Perform Addition and Subtraction operations from left to right.

Guided Practice

Evaluate the following expression for \begin{align*}x = 3, y = -2, z = -1\end{align*}.

\begin{align*}2z((x+1)^\frac{1}{2}-y^3)^2\end{align*}

Solution:

\begin{align*}2z((x+1)^\frac{1}{2}-y^3)^2&=2(-1)(((3)+1)^\frac{1}{2}-(-2)^3)^2\\ &=-2(4^\frac{1}{2}+8)^2\\ &=-2(2+8)^2\\ &=-2(10)^2\\ &=-200\end{align*}

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