Question

$2^x+3^{x+4}=7^x,\text{then}x$

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

Answer 1

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

We have,

$2^x+3^{x+4}=7^x$

On taking logarithm both sides, we get

$x\log 2+(x+4)\log 3=x\log 7$

$x\log 2+x\log 3+4\log 3=x\log 7$

$x(\log 2+\log 3)+4\log 3=x\log 7$

$x\log (3\times 2)+4\log 3=x\log 7$

$x\log 6+4\log 3=x\log 7$

$4\log 3=x(\log 7-\log 6)$

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

Hence, this is the answer.