Question Number 155583 by cortano last updated on 02/Oct/21
Answered by yeti123 last updated on 02/Oct/21
$${x}\:=\:\mathrm{6}.\mathrm{4172604685892}... \\ $$$${x}\:\approx\:\mathrm{6}.\mathrm{417260} \\ $$
Answered by peter frank last updated on 02/Oct/21
$$\mathrm{x}−\mathrm{5log}\:\left(\mathrm{5x}−\mathrm{2}\right)=\mathrm{2x}+\mathrm{1log}\:\left(\mathrm{x}−\mathrm{5}\right) \\ $$$$\left(\frac{\mathrm{x}−\mathrm{5}}{\mathrm{2x}+\mathrm{1}}\right)\mathrm{log}\:\left(\mathrm{5x}−\mathrm{2}\right)−\mathrm{log}\:\left(\mathrm{x}−\mathrm{5}\right)=\mathrm{0} \\ $$$$.... \\ $$$$ \\ $$