Question

Solve the following equation:
${\log }_7(2^x-1)+{\log }_7(2^x-7)=1$

Answer 1

${log}_7(2^x-1)+{log}_7(2^x-7)=1\Rightarrow {(2}^x-1\left)\right(2^x-7)=7\Rightarrow {{(2}^x)}^2-8(2^x)+7=7\Rightarrow {(2}^x)(2^x-8)=0\Rightarrow 2^x-8=0(2^x\ne 0)\Rightarrow 2^x=8\Rightarrow x=3$