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

Answer 1

The correct option is A $\frac{4{\log }_{e}3}{{\log }_{e}7-{\log }_{e}6}$

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

applying logarithm on both sides we get

${\log }_e\left(2^x{.3}^x{.3}^4\right)={\log }_e\left(7^x\right)$

$⟹{\log }_e\left(\right(2.3{)}^x{.3}^4)={\log }_e(7^x)$

$⟹{\log }_e\left(6^x\right)+{\log }_e\left(3^4\right)={\log }_e\left(7^x\right)$

$⟹x{\log }_{e}6+4{\log }_{e}3=x{\log }_{e}7$

$⟹x({\log }_{e}7-{\log }_{e}6)=4{\log }_{e}3$

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