Question

If 9^2 x 3^2 x 3^n - 27^n/3^3m x 2^3 = 1/27

Answer 1

Answer:

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Answer 2

ANSWER:

Given:

$\:\:\bullet\:\:\dfrac{9^n\times3^2\times3^n-27^n}{3^{3m}\times2^3}=\dfrac{1}{27}$

To Prove:

• m - n = 1

Proof:

We are given that,

$\implies\dfrac{9^n\times3^2\times3^n-27^n}{3^{3m}\times2^3}=\dfrac{1}{27}$

$\implies\dfrac{(3^2)^n\times3^2\times3^n-(3^3)^n}{3^{3m}\times2^3}=\dfrac{1}{3^3}$

$\implies\dfrac{3^{2n}\times3^2\times3^n-3^{3n}}{3^{3m}\times2^3}=\dfrac{1}{3^3}$

We know that,

$\hookrightarrow a^x\times a^y=a^{x+y}$

So,

$\implies\dfrac{3^{2n}\times3^2\times3^n-3^{3n}}{3^{3m}\times2^3}=\dfrac{1}{3^3}$

$\implies\dfrac{3^{2n+2+n}-3^{3n}}{3^{3m}\times2^3}=\dfrac{1}{3^3}$

So,

$\implies\dfrac{3^{3n+2}-3^{3n}}{3^{3m}\times2^3}=\dfrac{1}{3^3}$

Taking, 3^(3n) common in numerator,

$\implies\dfrac{3^{3n+2}-3^{3n}}{3^{3m}\times2^3}=\dfrac{1}{3^3}$

$\implies\dfrac{(3^{3n})(3^2-1)}{3^{3m}\times8}=\dfrac{1}{3^3}$

$\implies\dfrac{(3^{3n})(8\!\!\!/)}{3^{3m}\times8\!\!\!/}=\dfrac{1}{3^3}$

So,

$\implies\dfrac{3^{3n}}{3^{3m}}=\dfrac{1}{3^3}$

We know that,

$\hookrightarrow \dfrac{a^x}{a^y}=a^{x-y}$

And,

$\hookrightarrow \dfrac{1}{a^y}=a^{-y}$

So,

$\implies\dfrac{3^{3n}}{3^{3m}}=\dfrac{1}{3^3}$

$\implies3^{3n-3m}=3^{-3}$

As the bases are same, we compare the powers

$\implies3n-3m=-3$

$\implies3\!\!\!/(n-m)=-3\!\!\!/$

So,

$\implies n-m=-1$

$\implies -(m-n)=-1$

Hence,

$\implies\bf m-n=1$

HENCE PROVED!!