Question

Let $a,b\in R^{+}$ 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 $ab$ is

• A. $2^7$
• B. $7^2$
• C. $3^5$
• D. $5^3$

Answer 1

The correct option is C $3^5$

${\log }_{27}a+{\log }_{9}b=\frac{7}{2}$
$\Rightarrow \frac{1}{3}{\log }_{3}a+\frac{1}{2}{\log }_{3}b=\frac{7}{2}$
$\Rightarrow 2{\log }_{3}a+3{\log }_{3}b=21\cdots \left(1\right)$

${\log }_{27}b+{\log }_{9}a=\frac{2}{3}$
$\Rightarrow \frac{1}{2}{\log }_{3}a+\frac{1}{3}{\log }_{3}b=\frac{2}{3}$
$\Rightarrow 3{\log }_{3}a+2{\log }_{3}b=4\cdots \left(2\right)$

Adding equations $\left(1\right)$ and $\left(2\right)$, we get
$5{\log }_{3}a+5{\log }_{3}b=25$
$\Rightarrow 5{\log }_3\left(ab\right)=25$
$\Rightarrow {\log }_3\left(ab\right)=5$
$\therefore ab=3^5$