log-5-4x-log-10-x-find-x-

Question Number 97725 by john santu last updated on 09/Jun/20

$$\mathrm{log}\:_{\mathrm{5}} \left(\mathrm{4x}\right)=\mathrm{log}\:_{\mathrm{10}} \left(\mathrm{x}\right) \\ $$$$\mathrm{find}\:\mathrm{x}\:? \\ $$

Commented by prakash jain last updated on 09/Jun/20

4x=5^y x=10^y 4×10^y =5^y 2^y =(1/4)⇒y=−2 x=(1/(100))

$$\mathrm{4}{x}=\mathrm{5}^{{y}} \\ $$$${x}=\mathrm{10}^{{y}} \\ $$$$\mathrm{4}×\mathrm{10}^{{y}} =\mathrm{5}^{{y}} \\ $$$$\mathrm{2}^{{y}} =\frac{\mathrm{1}}{\mathrm{4}}\Rightarrow{y}=−\mathrm{2} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{100}} \\ $$

Commented by bobhans last updated on 09/Jun/20

((ln(4x))/(ln(5))) = ((ln(x))/(ln(10))) ((2ln(2)+ln(x))/(ln(5))) = ((ln(x))/(ln(5)+ln(2))) ln(5)ln(x)=(2ln(2)+ln(x))(ln(5)+ln(2)) ln(5)ln(x)=2ln(5)ln(2)+2ln^2 (2)+ln(5)ln(x)+ln(x)ln(2) 0 = 2ln(5)ln(2)+2ln^2 (2)+ln(x)ln(2) 0 = 2ln(5)+2ln(2)+ln(x) ln(x)=−2ln(10)=ln(10^(−2) ) ⇔x = (1/(100))

$$\frac{\mathrm{ln}\left(\mathrm{4x}\right)}{\mathrm{ln}\left(\mathrm{5}\right)}\:=\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{10}\right)} \\ $$$$\frac{\mathrm{2ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{5}\right)}\:=\:\frac{\mathrm{ln}\left(\mathrm{x}\right)}{\mathrm{ln}\left(\mathrm{5}\right)+\mathrm{ln}\left(\mathrm{2}\right)} \\ $$$$\mathrm{ln}\left(\mathrm{5}\right)\mathrm{ln}\left(\mathrm{x}\right)=\left(\mathrm{2ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{x}\right)\right)\left(\mathrm{ln}\left(\mathrm{5}\right)+\mathrm{ln}\left(\mathrm{2}\right)\right) \\ $$$$\mathrm{ln}\left(\mathrm{5}\right)\mathrm{ln}\left(\mathrm{x}\right)=\mathrm{2ln}\left(\mathrm{5}\right)\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{5}\right)\mathrm{ln}\left(\mathrm{x}\right)+\mathrm{ln}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{0}\:=\:\mathrm{2ln}\left(\mathrm{5}\right)\mathrm{ln}\left(\mathrm{2}\right)+\mathrm{2ln}^{\mathrm{2}} \left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{x}\right)\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{0}\:=\:\mathrm{2ln}\left(\mathrm{5}\right)+\mathrm{2ln}\left(\mathrm{2}\right)+\mathrm{ln}\left(\mathrm{x}\right) \\ $$$$\mathrm{ln}\left(\mathrm{x}\right)=−\mathrm{2ln}\left(\mathrm{10}\right)=\mathrm{ln}\left(\mathrm{10}^{−\mathrm{2}} \right) \\ $$$$\Leftrightarrow\mathrm{x}\:=\:\frac{\mathrm{1}}{\mathrm{100}} \\ $$

Commented by john santu last updated on 09/Jun/20

$$\mathrm{thank}\:\mathrm{both} \\ $$

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