Question

Solve for y if :
$\frac{{\left(\frac{1}{9}\right)}^{2y-1}(.0081{)}^{\frac{1}{3}}}{\sqrt{243}}={\left(\frac{1}{3}\right)}^{2y-5}\sqrt[3]{3\frac{{27}^{y-1}}{10000}}$

• A. $19\times 8$
• B. $\frac{-19}{8}$
• C. $\frac{18}{9}$
• D. $\frac{-19}{18}$

Answer 1

The correct option is D $\frac{-19}{18}$

Given:
$\frac{{\left(\frac{1}{9}\right)}^{2y-1}(.0081{)}^{\frac{1}{3}}}{\sqrt{243}}={\left(\frac{1}{3}\right)}^{2y-5}\sqrt[3]{3\frac{{27}^{y-1}}{10000}}$

$\Rightarrow \frac{(3^{-2}{)}^{2y-1}.(3^4.{10}^{-4}{)}^{\frac{1}{3}}}{3^{\frac{5}{2}}}$ = $\frac{3^{-(2y-5)}{.3}^{3\left(\frac{y-1}{3}\right)}}{{10}^{\frac{4}{3}}}$
$\Rightarrow \frac{3^{-4y+2+\frac{4}{3}-\frac{5}{2}}}{{10}^{\frac{4}{3}}}=\frac{3^{-2y+5+y-1}}{{10}^{\frac{4}{3}}}$
$\Rightarrow 3^{-4y+\frac{5}{6}}=3^{-y+4}$

On equating the exponents with same bases, we get
$\Rightarrow -4y+\frac{5}{6}=-y+4$
$\Rightarrow y=\frac{-19}{18}$