Question

If 9^n×3^2×3^n-(27)^n/(3^m×2^3)=3^-3 Show that:m-n=1

Answer 1

Hey!!

Your correct question:

If $\tt {\frac {9^{n} \times 3^{2} \times 3^{n} - 27^{n}}{3^{3m} \times 2^{3}} = 3^{-3}}$, show that : m - n = 1

Answer 2

Answer :

$\tt {\frac{9 {}^{n} \times 3 {}^{2} \times 3 {}^{n} - 27 {}^{n} }{3 {}^{3m} \times 2 {}^{ - 3}} } = {(3)}^{ - 3}$

$\tt{\rightarrow \frac{3 {}^{2n + 2 + n} - 3 {}^{3n} }{3 {}^{3m} \times 2 {}^{3}} = {(3)}^{ - 3}}$

$\tt{\rightarrow \frac{3 {}^{3n + 2} - 3 {}^{3n} }{3 {}^{3m} \times 2 {}^{3}} = {(3)}^{ - 3}}$

Taking 3^3n common in the numerator,

$\tt{ \rightarrow\frac{3 {}^{3n} (3 {}^{2} - 1)}{ {3}^{3m} \times 2 {}^{3} } = {(3)}^{ - 3}}$

$\tt{ \rightarrow\frac{3 {}^{3n} (9 - 1)}{ {3}^{3m} \times 8 } = {(3)}^{ - 3}}$

$\tt{\rightarrow \frac{3 {}^{3n} }{3 {}^{3m} } = {(3)}^{ - 3}}$

Therefore,

$\tt {(3) {}^{3n - 3m} = (3) {}^{ - 3}}$

$\tt{ \rightarrow3n - 3m = - 3}$

$\tt {\rightarrow n - m = - 1}$

$\huge \tt{\rightarrow m - n = 1}$

Hence, Proved.