Question

$\huge{ \mathbb{ \purple{HEYA \: \: FOLKS !!}}}$
Here is a question for Mathlovers !!
Solve for ' y ' :
$\ \textless \ br /\ \textgreater \ \sf\dfrac{\left(\dfrac{1}{9}\right)^{2y - 1}(0.0081)^{\frac{1}{3}}}{\sqrt{243}} = \left(\dfrac{1}{3}\right)^{2y - 5} \sqrt[3]{\dfrac{(27)^{y - 1}}{10000}} \\ \ \textless \ br /\ \textgreater \ \sf A)\frac{1}{2} \qquad \qquad B)\frac{-19}{18}\\\\\sf C)\frac{3}{10} \qquad \qquad D)\frac{12}{17}$
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Answer 1

AnswEr :

$\bf\dag\:\: \underline{Given :}\\\\ \sf\dfrac{\left(\dfrac{1}{9}\right)^{\normalsize2y - 1}(0.0081)^{\Large\frac{1}{3}}}{\sqrt{243}} = \left(\dfrac{1}{3}\right)^{\normalsize2y - 5} \sqrt[3]{\dfrac{(27)^{y - 1}}{10000}} \\\\\\\bf\dag\:\:\underline{To\:Find :}\\\\\textsf{The value of y.}$

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☢$\:\:\underline{\textsf{Let's Head to the Question Now :}}$

$:\implies\sf\dfrac{\left(\dfrac{1}{9}\right)^{\normalsize2y - 1}(0.0081)^{\large\frac{1}{3}}}{\sqrt{243}} = \left(\dfrac{1}{3}\right)^{\normalsize2y - 5} \sqrt[3]{\dfrac{(27)^{y - 1}}{10000}}\\\\\\: \implies\sf \dfrac{\left[(3)^{-2}\right]^{2y - 1}\left[(0.3)^4\right]^{\large\frac{1}{3}}}{\left[(3)^5\right]^{\large\frac{1}{2}}} =\left[(3)^{-1}\right]^{2y - 5} \left[\dfrac{\left[(3)^3\right]^{y - 1}}{(10)^4}\right]^{\large\frac{1}{3}}$

$:\implies\sf\dfrac{(3)^{-2(2y - 1)} \times (0.3)^{\large\frac{4}{3}}}{(3)^{\large\frac{5}{2}}} = (3)^{-(2y - 5)} \left[\dfrac{(3)^{3(y - 1)}}{(10)^4}\right]^{\large\frac{1}{3}}\\ \\\\:\implies\sf\dfrac{(3)^{-4y + 2} \times (0.3)^{\large\frac{4}{3}}}{(3)^{\large\frac{5}{2}}} = (3)^{-2y + 5} \left[\dfrac{(3)^{\large\frac{3(y - 1)}{3}}}{(10)^{\large\frac{4}{3}}}\right]$

$:\implies \sf\dfrac{(3)^{-4y + 2} \times \left(\dfrac{3}{10}\right)^{\large\frac{4}{3}}}{(3)^{\large\frac{5}{2}}} = (3)^{-2y + 5} \left[\dfrac{(3)^{(y - 1)}}{(10)^{\large\frac{4}{3}}}\right] \\\\\\:\implies\sf\dfrac{(3)^{-4y + 2} \times (3)^{\large\frac{4}{3}}\times \left(\dfrac{1}{10}\right)^{\large\frac{4}{3}}}{(3)^{\large\frac{5}{2}}} = \dfrac{(3)^{-2y + 5} \times (3)^{(y - 1)}}{(10)^{\large\frac{4}{3}}}$

$: \implies\sf(3)^{-4y + 2} \times (3)^{\large\frac{4}{3}} \times (3)^{\large\frac{-5}{2}} = \dfrac{(3)^{-2y + 5 + y - 1}}{\left(\dfrac{1}{10}\right)^{\large\frac{4}{3}} \times (10)^{\large\frac{4}{3}}}\\\\\\: \implies\sf(3)^{-4y + 2} \times (3)^{\bigg[\large\frac{4}{3} - \frac{5}{2}\bigg]} = (3)^{-y + 4}\\\\\\:\implies\sf(3)^{\bigg[\large\frac{8 - 15}{6}\bigg]} = \dfrac{(3)^{-y + 4}}{(3)^{-4y + 2}}$

$:\implies \sf(3)^{\bigg[\large\frac{ - 7}{6}\bigg]} = (3)^{-y + 4 - (-4y + 2)}\\\\\\:\implies \sf(3)^{\bigg[\large\frac{-7}{6}\bigg]} = (3)^{-y + 4 + 4y - 2}\\\\\\:\implies\sf(3)^{\bigg[ \large\frac{-7}{6}\bigg]} = (3)^{3y + 2}\\\\{\scriptsize\qquad\bf{\dag}\:\texttt{When Bases are same Exponents should be Equal}}\\\\:\implies \sf3y + 2 = \dfrac{-7}{6}\\\\\\:\implies \sf6(3y + 2) = - 7\\\\\\:\implies \sf18y + 12 = - 7\\\\\\:\implies \sf18y = - 7 - 12\\\\\\:\implies \sf18y = -19\\\\\\:\implies \underline{\boxed{\large\red{\sf y = \dfrac{-19}{18}}}}$

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$\boxed{\begin{minipage}{5 cm}\bf{\dag}\:\:\underline{\text{Law of Exponents :}}\\\\\bigstar\:\:\sf\dfrac{a^m}{a^n} = a^{m - n}\\\\\bigstar\:\:\sf{(a^m)^n = a^{mn}}\\\\\bigstar\:\:\sf(a^m)(a^n) = a^{m + n}\\\\\bigstar\:\:\sf\dfrac{1}{a^n} = a^{-n}\\\\\bigstar\:\:\sf\sqrt[\sf n]{\sf a} = (a)^{\dfrac{1}{n}}\end{minipage}}$