DAV Class 8 Maths Chapter 3 Enrichment Question
Exponents and Radicals Enrichment Question
1. \( 9^x \times 3^2 \times (3^{\frac{-x}{2}})^{-2} = \dfrac{1}{27}, \) find \( x \)
Solution
\[ \begin{aligned} 9^x \times 3^2 \times (3^{\frac{-x}{2}})^{-2} &= \frac{1}{27} \\[8pt] (3^2)^x \times 3^2 \times 3^{\left(\frac{-x}{\cancel2_1} \times -{\cancel2^1} \right)} &= \frac{1}{3^3} \\[8pt] 3^{2x} \times 3^2 \times 3^x &= 3^{-3} \\[8pt] 3^{2x + 2+ x} &= 3^{-3} \\[8pt] \text{Bases are the same, } & \text{we equate powers.} \\[8pt] 2x + 2+ x & = -3 \\[8pt] 3x & = -3 -2 \\[8pt] 3x & = -5 \\[8pt] x & = \frac{-5}{3} \end{aligned} \]
Answer \( x = \color{red} \dfrac{-5}{3} \)
2. By what number should \( \left(\dfrac{-3}{2}\right)^{-3} \) be divided so that the quotient may be \( \left(\dfrac{-8}{27}\right)^{-2} \)
Solution
\[ \begin{aligned} \text{Let the number be } & x \\ \left(\frac{-3}{2}\right)^{-3} \div x & = \left(\frac{-8}{27}\right)^{-2} \\[8pt] \left(\frac{-3}{2}\right)^{-3} & = \left(\frac{-8}{27}\right)^{-2} \times x \\[8pt] \left(\frac{-3}{2}\right)^{-3} \div \left(\frac{-8}{27}\right)^{-2} & = x \\[8pt] \left(\frac{-2}{3}\right)^{3} \div \left(\frac{-27}{8}\right)^{2} & = x \\[8pt] \left(\frac{-2}{3}\right)^{3} \times \left(\frac{-2^3}{3^3}\right)^{2} & = x \\[8pt] \left(\frac{-2}{3}\right)^{3} \times \left(\frac{-2}{3}\right)^{3 \times 2} & = x \\[8pt] \left(\frac{-2}{3}\right)^{3} \times \left(\frac{-2}{3}\right)^{6} & = x \\[8pt] \left(\frac{-2}{3}\right)^{9} & = x \end{aligned} \]
Answer \( x = \color{red} \left(\dfrac{-2}{3}\right)^{9} \)