Question Number 49463 by Pk1167156@gmail.com last updated on 07/Dec/18
$$\mathrm{If}\:\:\:\mathrm{7}^{\mathrm{log}\:_{\mathrm{7}} \left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)} =\:{x}−\mathrm{1},\:\:{x}\:\mathrm{may}\:\mathrm{have} \\ $$$$\mathrm{values} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$${a}^{{x}} ={m}\:\:\:\:{x}={log}_{{a}} {m} \\ $$$${log}_{\mathrm{7}} \left({x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}\right)={log}_{\mathrm{7}} \left({x}−\mathrm{1}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}={x}−\mathrm{1} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2}\:\:{and}\:\mathrm{3} \\ $$