Question

Simplify the following :

${\left(\frac{2}{3}\right)}^4\times 3^2\times {\left(\frac{-1}{2}\right)}^5$

Answer 1

powers and exponent

Step 1: Find the value of ${\left(\frac{2}{3}\right)}^4$, $3^2$ and ${\left(\frac{-1}{2}\right)}^5$.

Given, ${\left(\frac{2}{3}\right)}^4\times 3^2\times {\left(\frac{-1}{2}\right)}^5$

So,

$$\begin{gathered} {\left(\frac{2}{3}\right)}^4=\left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right)\times \left(\frac{2}{3}\right) \\ =\frac{2\times 2\times 2\times 2}{3\times 3\times 3\times 3} \\ =\frac{16}{81} \end{gathered}$$

${\left(3\right)}^2=3\times 3=9$

$$\begin{gathered} {\left(\frac{-1}{2}\right)}^5=\left(\frac{-1}{2}\right)\times \left(\frac{-1}{2}\right)\times \left(\frac{-1}{2}\right)\times \left(\frac{-1}{2}\right)\times \left(\frac{-1}{2}\right) \\ =-\frac{1}{32} \end{gathered}$$

Step 2: Find the value of ${\left(\frac{2}{3}\right)}^4\times 3^2\times {\left(\frac{-1}{2}\right)}^5$

Now,

$$\begin{gathered} {\left(\frac{2}{3}\right)}^4\times 3^2\times {\left(\frac{-1}{2}\right)}^5=\frac{16}{81}\times 9\times \left(\frac{-1}{32}\right) \\ =\frac{{\cancel{16}}^1}{{\cancel{81}}^9}\times \cancel{9}\times \frac{-1}{{\cancel{32}}^2} \\ =\frac{1}{9}\times 1\times \frac{-1}{2} \\ =-\frac{1}{18} \end{gathered}$$

Hence, ${\left(\frac{2}{3}\right)}^4\times 3^2\times {\left(\frac{-1}{2}\right)}^5=-\frac{1}{18}$.