Question

Suppose that $a$ and $b$ are 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 find the value of $ab$

Answer 1

Given ${\log }_{27}a+{\log }_{9}b=\frac{7}{2}$

$\Rightarrow {\log }_{3^3}a+{\log }_{3^2}b=\frac{7}{2}$

$\Rightarrow \frac{1}{3}{\log }_{3}a+\frac{1}{2}{\log }_{3}b=\frac{7}{2}$ ...(1) ($\because {\log }_{a^b}x=\frac{1}{b}{\log }_{a}x$)

Also given ${\log }_{27}b+{\log }_{9}a=\frac{2}{3}$

$\Rightarrow {\log }_{3^3}b+{\log }_{3^2}a=\frac{2}{3}$

$\Rightarrow \frac{1}{3}{\log }_{3}b+\frac{1}{2}{\log }_{3}a=\frac{2}{3}$ ...(2) ($\because {\log }_{a^b}x=\frac{1}{b}{\log }_{a}x$)

Solving (1) and (2), we get
${\log }_{3}a=-6$

Converting into exponential form

$\Rightarrow a=3^{-6}$

Put this value in (1), we get

$\frac{1}{2}{\log }_{3}b=\frac{11}{2}$

$\Rightarrow {\log }_{3}b=11$

Converting into exponential form

$\Rightarrow b=3^{11}$

$\therefore ab=3^{11}\times 3^{-6}=3^5$