Question

81^2/729^1-x=9^2x. answer me plz​

Answer 1

Answer:

The value of x is - 1.

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\dfrac{81^2}{729^{1\:-\:x}} \:=\:9^{2x}}$

We have to find the value of x.

Now,

$\displaystyle{\sf\:\dfrac{81^2}{729^{1\:-\:x}} \:=\:9^{2x}}$

$\displaystyle{\implies\sf\:\dfrac{(\:9^2\:)^2}{(\:9^3\:)^{1\:-\:x}}\:=\:9^{2x}}$

We know that,

$\displaystyle{\boxed{\pink{\sf\:(\:a^m\:)^n\:=\:a^{m\:\times\:n}\:}}}$

$\displaystyle{\implies\sf\:\dfrac{9^{2\:\times\:2}}{9^{3\:(\:1\:-\:x\:)}}\:=\:9^{2x}}$

$\displaystyle{\implies\sf\:\dfrac{9^4}{9^{3\:-\:3x}}\:=\:9^{2x}}$

$\displaystyle{\implies\sf\:9^{2x}\:\times\:9^{3\:-\:3x}\:=\:9^4}$

We know that,

$\displaystyle{\boxed{\blue{\sf\:a^m\:\times\:a^n\:=\:a^{m\:+\:n}\:}}}$

$\displaystyle{\implies\sf\:9^{2x\:+\:(\:3\:-\:3x\:)}\:=\:9^4}$

As bases are equal, powers must also be equal.

$\displaystyle{\therefore\:\sf\:2x\:+\:(\:3\:-\:3x\:)\:=\:4}$

$\displaystyle{\implies\sf\:2x\:+\:3\:-\:3x\:=\:4}$

$\displaystyle{\implies\sf\:2x\:-\:3x\:=\:4\:-\:3}$

$\displaystyle{\implies\sf\:-\:x\:=\:1}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:x\:=\:-\:1\:}}}}$

∴ The value of x is - 1.