Question

Solve: $\log 7^x=\log 9^{x-2}$

Answer 1

$\log 7^x=\log 9^{x-2}[\because \log a^x=x\log a]$

$x\log 7=(x-2)\log 9$

$x\log 7=x\log 9-2\log 9$

$x(\log 7-\log 9)=-2\log 9$

$x=\frac{\log 81}{\log 9-\log 7}$