Question

Suppose that $a$ and $b$ are two positive real numbers such that ${\log }_{27}a+{\log }_{9}b=\frac{7}{2}$ and ${\log }_{27}b+{\log }_{9}a=\frac{2}{3}$, then the value of product $ab$ is

Answer 1

${\log }_{27}a+{\log }_{9}b=\frac{7}{2}$
$\Rightarrow {\log }_{3^3}a+{\log }_{3^2}b=\frac{7}{2}$
$\Rightarrow \frac{1}{3}{\log }_{3}a+\frac{1}{2}{\log }_{3}b=\frac{7}{2}$
$\Rightarrow {\log }_3{a}^{1/3}+{\log }_3{b}^{1/2}=\frac{7}{2}$
$\Rightarrow {\log }_3\left(a^{1/3}{b}^{1/2}\right)=\frac{7}{2}$
$\Rightarrow a^{1/3}{b}^{1/2}=3^{7/2}...\left(1\right)$

${\log }_{27}b+{\log }_{9}a=\frac{2}{3}$
$\Rightarrow {\log }_{3^3}b+{\log }_{3^2}a=\frac{2}{3}$
$\Rightarrow \frac{1}{3}{\log }_{3}b+\frac{1}{2}{\log }_{3}a=\frac{2}{3}$
$\Rightarrow {\log }_3{b}^{1/3}+{\log }_3{a}^{1/2}=\frac{2}{3}$
$\Rightarrow {\log }_3\left(b^{1/3}{a}^{1/2}\right)=\frac{2}{3}$
$\Rightarrow a^{1/2}{b}^{1/3}=3^{2/3}...\left(2\right)$

Multiplying eqn $\left(1\right)$ and $\left(2\right),$ we get
$a^{5/6}{b}^{5/6}=3^{25/6}$
$\Rightarrow (ab{)}^{5/6}=3^{25/6}$
$\Rightarrow ab=3^5=243$