Question Number 212976 by hardmath last updated on 27/Oct/24
$$\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:=\:\mathrm{125}\:\:\:\mathrm{and}\:\:\:\mathrm{8}^{\boldsymbol{\mathrm{y}}} \:=\:\mathrm{5} \\ $$$$\mathrm{find}:\:\:\:\frac{\mathrm{2x}\:−\:\mathrm{y}}{\mathrm{y}}\:=\:? \\ $$
Answered by Amiltonsad last updated on 27/Oct/24
$$ \\ $$$${how}\:\mathrm{4}^{{x}\:} =\mathrm{125}\:\:\Rightarrow\:\mathrm{2}^{\mathrm{2}{x}} \:=\:\mathrm{5}^{\mathrm{3}} \:{and}\:\left(\mathrm{2}\right)^{\frac{\mathrm{2}{x}}{\mathrm{3}}} \:=\:\mathrm{5} \\ $$$${and}\:{how}…\:\mathrm{8}^{{y}} \:=\:\mathrm{5}\:\:\mathrm{2}^{\mathrm{3}{y}} \:=\:\mathrm{5} \\ $$$$ \\ $$$${finally}\:{we}\:{can}\:{say}\:\frac{\mathrm{2}{x}\:−\:{y}}{{y}}\:=\:\frac{\mathrm{9}{y}\:−\:{y}}{{y}}\:=\:\frac{\mathrm{8}{y}}{{y}}\:=\mathrm{8} \\ $$$${answered}\:{by}\:{Alfa}\:{dos}\:{ALFAS} \\ $$
Answered by A5T last updated on 27/Oct/24
$${x}={log}_{\mathrm{4}} \mathrm{125}=\frac{\mathrm{3}}{\mathrm{2}}{log}_{\mathrm{2}} \mathrm{5};\:{y}={log}_{\mathrm{8}} \mathrm{5}=\frac{{log}_{\mathrm{2}} \mathrm{5}}{\mathrm{3}} \\ $$$$\Rightarrow\frac{\mathrm{2}{x}−{y}}{{y}}=\frac{\mathrm{2}{x}}{{y}}−\mathrm{1}=\frac{\mathrm{3}{log}_{\mathrm{2}} \mathrm{5}}{\frac{{log}_{\mathrm{2}} \mathrm{5}}{\mathrm{3}}}−\mathrm{1}=\mathrm{8} \\ $$