Question Number 84997 by M±th+et£s last updated on 18/Mar/20
$${log}_{\frac{{x}}{\mathrm{2}}} {x}^{\mathrm{2}} −{log}_{\mathrm{16}{x}} {x}^{\mathrm{3}} +\mathrm{40}{log}_{\mathrm{4}{x}} \sqrt{{x}}=\mathrm{0} \\ $$
Commented by jagoll last updated on 18/Mar/20
![((log_2 (x^2 ))/(log_2 (x)−1))−((log_2 (x^3 ))/(log_2 (x)+4)) + ((40 log_2 ((√x)))/(log_2 (x)+2)) = 0 let log_2 (x) = t ((2t)/(t−1))−((3t)/(t+4))+((20t)/(t+2)) = 0 t [ (2/(t−1))−(3/(t+4)) + ((20)/(t+2)) ] = 0](https://www.tinkutara.com/question/Q84999.png)
$$\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}^{\mathrm{2}} \right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{1}}−\frac{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}^{\mathrm{3}} \right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{4}}\:+\:\frac{\mathrm{40}\:\mathrm{log}_{\mathrm{2}} \:\left(\sqrt{\mathrm{x}}\right)}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{2}}\:=\:\mathrm{0} \\ $$$$\mathrm{let}\:\mathrm{log}_{\mathrm{2}} \:\left(\mathrm{x}\right)\:=\:\mathrm{t} \\ $$$$\frac{\mathrm{2t}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{3t}}{\mathrm{t}+\mathrm{4}}+\frac{\mathrm{20t}}{\mathrm{t}+\mathrm{2}}\:=\:\mathrm{0} \\ $$$$\mathrm{t}\:\left[\:\frac{\mathrm{2}}{\mathrm{t}−\mathrm{1}}−\frac{\mathrm{3}}{\mathrm{t}+\mathrm{4}}\:+\:\frac{\mathrm{20}}{\mathrm{t}+\mathrm{2}}\:\right]\:=\:\mathrm{0} \\ $$
Commented by jagoll last updated on 18/Mar/20
$$\mathrm{now}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve} \\ $$