Question

If $\frac{9^b\times 3^2\times {\left(3^{\frac{-b}{2}}\right)}^{-2}-{\left(27\right)}^b}{3^{3a}\times 2^3}=\frac{1}{27}$, then prove that $a-b=1$.

Answer 1

$\frac{9^b\times 3^2\times \left(3^{{\frac{-b}{2}}^2}\right)-{27}^b}{3^{3a}\times 2^3}=\frac{1}{27}\Rightarrow \frac{3^{2b}\times 3^2\times \left(3^b\right)-3^{3b}}{3^{3a}\times 2^3}=\frac{1}{27}\Rightarrow \frac{3^{3b}\times 3^2-3^{3b}}{3^{3a}\times 2^3}=\frac{1}{3^3}\Rightarrow \frac{3^{3b}(9-1)}{3^{3a}\times 8}=\frac{1}{3^3}$

$\Rightarrow \frac{3^{3b}}{3^{3a}}=\frac{1}{3^3}$

Taking ${\log }_3$ both sides,

$\Rightarrow {\log }_3\left(\frac{3^{3b}}{3^{3a}}\right)={\log }_3\left(3^{-3}\right)$

$\Rightarrow {\log }_3{3}^{3b}-{\log }_3{3}^{3a}={\log }_3\left(3^{-3}\right)$

$\Rightarrow 3b-3a=-3$

$\Rightarrow a-b=1$