Question

Can anyone solve this?????​

Answer 1

$\mathsf{Given :\;\dfrac{\left(\dfrac{1}{9}\right)^{2y - 1}(0.0081)^{\dfrac{1}{3}}}{\sqrt{243}} = \left(\dfrac{1}{3}\right)^{2y - 5} \sqrt[3]{\dfrac{(27)^{y - 1}}{10000}}}$

$\mathsf{\implies \dfrac{\left(\dfrac{1}{9}\right)^{2y - 1}(0.0081)^{\dfrac{1}{3}}}{(243)^{\dfrac{1}{2}}} = \left(\dfrac{1}{3}\right)^{2y - 5} \left[\dfrac{(27)^{y - 1}}{10000}\right]^{\dfrac{1}{3}}}$

Consider the terms in LHS :

$\bigstar\;\;\boxed{\mathsf{\dfrac{1}{9} = \dfrac{1^2}{3^2} = \left(\dfrac{1}{3}\right)^2 = \left[(3)^{-1}\right]^2 = 3^{-2}}}$

$\bigstar\;\;\boxed{\mathsf{0.0081 = (0.09)^2 = \left[(0.3)^2\right]^2 = (0.3)^4}}$

$\bigstar\;\;\boxed{\mathsf{243 = (3)^5}}$

Consider the terms in RHS :

$\bigstar\;\;\boxed{\mathsf{\dfrac{1}{3} = (3)^{-1}}}$

$\bigstar\;\;\boxed{\mathsf{27 = (3)^3}}$

$\bigstar\;\;\boxed{\mathsf{10000 = (10)^4}}$

$\implies \mathsf{\dfrac{\left[(3)^{-2}\right]^{2y - 1}\left[(0.3)^4\right]^{\dfrac{1}{3}}}{\left[(3)^5\right]^{\dfrac{1}{2}}} = \left[(3)^{-1}\right]^{2y - 5} \left[\dfrac{\left[(3)^3\right]^{y - 1}}{(10)^4}\right]^{\dfrac{1}{3}}}}$

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

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

$\bigstar\;\;\boxed{\mathsf{0.3 = \dfrac{3}{10}}}$

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

$\implies \mathsf{\dfrac{(3)^{-4y + 2} \times (3)^{\dfrac{4}{3}} \times \left(\dfrac{1}{10}\right)^{\dfrac{4}{3}}}{(3)^{\dfrac{5}{2}}} = \dfrac{(3)^{-2y + 5} \times (3)^{(y - 1)}}{(10)^{\dfrac{4}{3}}}\right]}}$

$\implies \mathsf{(3)^{-4y + 2} \times (3)^{\dfrac{4}{3}} \times (3)^{\dfrac{-5}{2}} = \dfrac{(3)^{-2y + 5 + y - 1}}{\left(\dfrac{1}{10}\right)^{\dfrac{4}{3}} \times (10)^{\dfrac{4}{3}}}\right]}}$

$\implies \mathsf{(3)^{-4y + 2} \times (3)^{\bigg[\dfrac{4}{3} - \dfrac{5}{2}\bigg]} = (3)^{-y + 4}}$

$\implies \mathsf{(3)^{\bigg[\dfrac{8 - 15}{6}\bigg]} = \dfrac{(3)^{-y + 4}}{(3)^{-4y + 2}}}$

$\implies \mathsf{(3)^{\bigg[\dfrac{8 - 15}{6}\bigg]} = (3)^{-y + 4 - (-4y + 2)}}$

$\implies \mathsf{(3)^{\bigg[\dfrac{-7}{6}\bigg]} = (3)^{-y + 4 + 4y - 2}}$

$\implies \mathsf{(3)^{\bigg[\dfrac{-7}{6}\bigg]} = (3)^{3y + 2}}$

★  When Bases are same - Exponents should be Equal

$\implies \mathsf{3y + 2 = \dfrac{-7}{6}}$

$\implies \mathsf{6(3y + 2) = - 7}$

$\implies \mathsf{18y + 12 = - 7}$

$\implies \mathsf{18y = - 7 - 12}$

$\implies \mathsf{18y = -19}$

$\implies \mathsf{y = \dfrac{-19}{18}}$

Law of Exponents used to solve the Problem :

$\bigstar\;\;\large\boxed{\mathsf{\dfrac{a^m}{a^n} = a^{m - n}}}$

$\bigstar\;\;\large\boxed{\mathsf{(a^m)^n = a^{mn}}}$

$\bigstar\;\;\large\boxed{\mathsf{(a^m)(a^n) = a^{m + n}}}$

$\bigstar\;\;\large\boxed{\mathsf{\dfrac{1}{a^n} = a^{-n}}}$

$\bigstar\;\;\large\boxed{\mathsf{\sqrt[n]{a} = (a)^{\dfrac{1}{n}}}}$