Question

If $2^x.3^{x+4}=7^x$, then $x$ is equal to

• A. $\frac{4{\log }_{e}3}{{\log }_{e}7-{\log }_{e}6}$
• B. $\frac{4{\log }_{e}3}{{\log }_{e}6-{\log }_{e}7}$
• C. $\frac{2{\log }_{e}3}{{\log }_{e}7-{\log }_{e}6}$
• D. $\frac{2{\log }_{e}3}{{\log }_{e}6-{\log }_{e}7}$

Answer 1

The correct option is A

$\frac{4{\log }_{e}3}{{\log }_{e}7-{\log }_{e}6}$

Explanation for the correct option:

Step 1. Find the value of $x$:

Given, $2^x.3^{x+4}=7^x$

$\Rightarrow$ $2^x.3^x.3^4=7^x$

$\Rightarrow$ $6^x.3^4=7^x$

$\Rightarrow$ $3^4={\left(\frac{7}{6}\right)}^x$

Step 2. Take log on both sides, we get

$\begin{array}{rcl}4{\log }_{e}3& =& x{\log }_e\left(\frac{7}{6}\right)\\ & =& x({\log }_{e}7–{\log }_{e}6)\end{array}$

$\mathbf{\therefore }\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{4}\mathbf{}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{e}}\mathbf{3}}{\mathbf{(}\mathbf{}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{e}}\mathbf{7}\mathbf{}\mathbf{–}\mathbf{}\mathbf{l}\mathbf{o}{\mathbf{g}}_{\mathbf{e}}\mathbf{6}\mathbf{)}}$

Hence, Option ‘A’ is Correct.