Question

calculate the missing value of "X" in fallowing expression :-


[11/9]^3 ×[9/11]^6. =[11/9]^2x-1


please do not spem... it's urgent..​

Answer 1

Answer :-

x=-1

Explanation :-

$\implies \bigg(\dfrac{11}{9}\bigg)^{3} \times\bigg(\dfrac{9}{11} \bigg)^{6}= \bigg(\dfrac{11}{9}\bigg)^{2x-1}$

$\implies \bigg(\dfrac{11}{3^{2}}\bigg)^{3} \times \bigg(\dfrac{3^{2}}{11} \bigg)^{6}=\bigg(\dfrac{11}{3^{2}}\bigg)^{2x-1}$

$\implies \dfrac{11^{3}}{3^{2\times 3}}\times \dfrac{3^{2\times6}}{11^6} =\dfrac{11^{2x-1}}{(3^{2})^{2x-1}}$

$\implies \dfrac{11^{3}}{3^{6}}\times\dfrac{3^{12}}{11^{6}}=\dfrac{11^{2x-1}}{(3^{2})2x-1}}$

$\implies \dfrac{3^{12-6}}{11^{6-3}}=\dfrac{11^{2x-1}}{(3^{2})^{2x-1}}$

$\implies \dfrac{3^{6}}{11^{3}} =\dfrac{11^{2x-1}}{(3^{2})^{2x-1}}$

$\implies \dfrac{3^{6}}{11^{3}} =\dfrac {(3^{-2})^{2x-1}} {11^{-2x-1}}$

$\implies \bigg(\dfrac{3^{2}}{11} \bigg)^{3}=\bigg(\dfrac{3^{2}}{11} \bigg)^{-(2x-1)$

$\implies 3=-(2x-1)$

$\implies 3+(-1)=(-2x+1)+(-1)$

$\implies 2=(-2x+1)-1$

$\implies 2=-2x+1-1$

$\implies 2=-2x$

$\implies\dfrac{2x}{2}=-\dfrac{2}{2}$

$\boxed{x=-1}$