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Exponents and Powers — MATHS STD 8 — Question

Gujarat BoardEnglish MediumSTD 8MATHSExponents and Powers2 Marks
Question
Solve the following
(i) $\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{2}\right)^{-3}$
(ii) $\left(\frac{2}{3}\right)^{-2} \times\left(\frac{2}{3}\right)^5$
✓

Answer

(i)
$
\begin{aligned}
\left(\frac{1}{2}\right)^{-2}+\left(\frac{1}{2}\right)^{-3} & =\left(\frac{1}{2}\right)^{-2} \times\left(\frac{1}{2}\right)^3 =\left(\frac{1}{2}\right)^{-2+3}=\left(\frac{1}{2}\right)^1=\frac{1}{2}
\end{aligned}
$
(ii)
$
\begin{aligned}
\left(\frac{2}{3}\right)^{-2} \times\left(\frac{2}{3}\right)^5 & =\left(\frac{2}{3}\right)^{-2+5}=\left(\frac{2}{3}\right)^3=\frac{(2)^3}{(3)^3} \\
& =\frac{2 \times 2 \times 2}{3 \times 3 \times 3}=\frac{8}{27}
\end{aligned}
$

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