Question

If $7^x=3^{{\log }_{9}7}\cdot 5^{{\log }_{25}49}$, then the value of $x$ is

• A. $\frac{1}{2}$
• B. $1$
• C. $\frac{3}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{3}{2}$

$7^x=3^{{\log }_{9}7}\cdot 5^{{\log }_{25}49}$
$\Rightarrow 7^x=3^{{\log }_{3^2}7}\cdot 5^{{\log }_{5^2}49}$
$\Rightarrow 7^x=3^{\frac{1}{2}{\log }_{3}7}\cdot 5^{\frac{1}{2}{\log }_{5}49}$
$\Rightarrow 7^x={\left(3^{{\log }_{3}7}\right)}^{1/2}\cdot {\left(5^{{\log }_{5}49}\right)}^{1/2}$
$\Rightarrow 7^x=(7{)}^{1/2}\cdot (49{)}^{1/2}(\because a^{{\log }_{a}m}=m)$
$\Rightarrow 7^x=(7{)}^{1/2}\cdot 7=7^{3/2}$
$\Rightarrow x=\frac{3}{2}$