Question Number 106024 by john santu last updated on 02/Aug/20
$$\mathrm{log}\:_{\mathrm{4}} \left(\mathrm{5x}−\mathrm{6}\right).\mathrm{log}\:_{\mathrm{x}} \left(\mathrm{256}\right)=\mathrm{8} \\ $$
Commented by Dwaipayan Shikari last updated on 02/Aug/20
$$\left\{\mathrm{2},\mathrm{3}\right\} \\ $$
Answered by Dwaipayan Shikari last updated on 02/Aug/20
$$\frac{{log}\left(\mathrm{5}{x}−\mathrm{6}\right).{log}\left(\mathrm{256}\right)}{{log}\left(\mathrm{4}\right).{log}\left({x}\right)}=\mathrm{8} \\ $$$$\frac{{log}\left(\mathrm{5}{x}−\mathrm{6}\right).\mathrm{8}{log}\left(\mathrm{2}\right)}{\mathrm{2}{log}\left(\mathrm{2}\right){log}\left({x}\right)}=\mathrm{8} \\ $$$$\frac{{log}\left(\mathrm{5}{x}−\mathrm{6}\right)}{{log}\left({x}\right)}=\mathrm{2} \\ $$$$\mathrm{5}{x}−\mathrm{6}={e}^{\mathrm{2}{log}\left({x}\right)} \\ $$$$\mathrm{5}{x}−\mathrm{6}={x}^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$$\begin{cases}{{x}=\mathrm{2}}\\{{x}=\mathrm{3}}\end{cases} \\ $$
Answered by bemath last updated on 02/Aug/20
$$\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{8}\:\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}{x}−\mathrm{6}\right).\mathrm{log}\:_{{x}} \left(\mathrm{2}\right)=\mathrm{8} \\ $$$$\mathrm{log}\:_{\mathrm{2}} \left(\mathrm{5}{x}−\mathrm{6}\right)=\mathrm{2}.\mathrm{log}\:_{\mathrm{2}} \left({x}\right) \\ $$$${x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0} \\ $$$${x}\:=\:\mathrm{2}\:\&\:{x}\:=\:\mathrm{3}\: \\ $$