Q. 1 - Precalculus Enhanced with Graphing Utilities

Question

$4^3 = \_\_\_\_\_\_\_ ;8^{2/3}= \_\_\_\_\_\_\_ ;3^{-2}= \_\_\_\_\_\_\_$

Step-by-Step Solution

Answer

The answer of  $$\begin{gathered} 4^3 = 64 \\ {8}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=4 \\ {3}^{-2}=\frac{1}{9} \end{gathered}$$.

Step 1. Solving the given expression

We can write $4^3$ as $4\times 4\times 4$.

The solution is $4\times 4\times 4=64.$

Step 2. Solving the given expression

By the laws of exponent, $a^{\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$n$}\right.}=\sqrt[n]{a^m}$.

So,

$$\begin{gathered} 8^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} =\sqrt[3]{8^2} \\ =\sqrt[3]{64} \\ =\sqrt[3]{{\left(4\right)}^3} \\ =4 \end{gathered}$$

Step 3. Solving the given expression

By the laws of exponent, $a^{-x}=\frac{1}{a^x}$.

So,

$$\begin{gathered} 3^{-2}=\frac{1}{3^2} \\ =\frac{1}{9} \end{gathered}$$