Question

If $800=8\times {10}^8\times x^{-3/2}$, then $x=$

• A. ${10}^2$
• B. ${10}^3$
• C. ${10}^4$
• D. ${10}^5$

Answer 1

Answer: (C)

$800=8\times {10}^8\times x^{-3/2}$

$\Rightarrow$ $\frac{800}{8\times {10}^8}=x^{-3/2}$

$\Rightarrow$ $\frac{100}{{10}^8}=\frac{1}{\sqrt{x^3}}$

$\Rightarrow$ $\frac{1}{{10}^6}=\frac{1}{\sqrt{x^3}}$

$\Rightarrow$ $\frac{1}{{10}^{12}}=\frac{1}{x^3}$ [Taking square on both sides]

$\Rightarrow$ $\frac{1}{{10}^4}=\frac{1}{x}$ [Taking cube root on both sides]

$\therefore$ $x={10}^4$