Question

Solve for $x:{11}^{4x-5}\cdot 3^{2x}=5^{3-x}\cdot 7^{-x}$

Answer 1

$\left({11}^{4x-5}\right){.3}^{2x}=5^{3-2x}{.7}^{-x}$
Taking $\log$ on both sides give us
$\log \left({11}^{4x-5}{.3}^{2x}\right)=\log \left(5^{3-2x}{7}^{-x}\right)$
$\log \left({11}^{4x-5}\right)+\log \left(3^{2x}\right)=\log \left(5^{3-2x}\right)+\log \left(7^{-x}\right)$
$(4x-5)\log 11+2x\log 3=(3-2x)\log 5-x\log 7$
$4x\log 11+2x\log 3-5\log 11=3\log 5-2x\log 5-x\log 7$
$2x(2\log 11++\log 3)-5\log 11=3\log 5-x(2\log 5+\log 7)$
$2x(\log 121+\log 3)=5\log 11+3\log 5-x(\log 25+\log 7)$
$2x\log 363+x\log \left(175\right)=5\log 11+3\log 5$
$x(2\log 363+\log 175)=5\log 11+3\log 5$
$x=\frac{5\log 11+3\log 5}{2\log 363+\log 175}$
Hence the value of $x$ satisfying the above equation is,
$x=\frac{5\log 11+3\log 5}{2\log 363+\log 175}$