Question

If $\frac{9^n\times 3^2\times 3^n-(27{)}^n}{\left(3^3\right)\times 2^3}=\frac{1}{27}$, find the value of n.

Answer 1

We have

$\frac{9^n\times 3^2\times 3^n-(27{)}^n}{(3^3{)}^5\times 2^3}=\frac{1}{27}$
$=\frac{(3^2{)}^n\times 3^2\times 3^n-(3^3{)}^n}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{2n+2+n}-(3{)}^{3n}}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n+2}-(3{)}^{3n}}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n}\times (3{)}^2-(3{)}^{3n}}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{\left(3{)}^{3n}\right(3^2-1)}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n}\times 8}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{(3{)}^{3n}\times 2^3}{(3{)}^{15}\times 2^3}=\frac{1}{27}$
$=\frac{3^{3n}}{3^{15}}=\frac{1}{27}$
$=3^{3n-15}=\frac{1}{3^3}$
$=3^{3n-15}=3^{-3}$
On equating the coefficients, we get
3n -15 = -3
⇒ 3n = -3 + 15
⇒ 3n = 12
⇒ n =$\frac{12}{3}=4$