Question

Solve the following equation:
${\log }_6{2}^{x+3}-{\log }_6(3^x-2)=x$

Answer 1

${{\log }_{6}2}^{x+3}-{\log }_{6}3x-2=x\frac{2^{x+3}}{3^x-2}=6^x{2}^x=a,3^x=b\frac{8a}{b-2}=ab8a=a(b^2-2b)a(b^2-2b-8)=0$

$a=0$ (not possible)

Therefore $2^x$ never be $0$

Therefore, $b^2-2b-8=0b=4,-23^x=4,-2$

$3^x=4$

$x={\log }_{3}4$