Question Number 146459 by mathdanisur last updated on 13/Jul/21
$${if}\:\:\:{x}^{\mathrm{2}} ={y}^{\mathrm{3}} \:\:\:{find}\:\:\:\:{log}_{\boldsymbol{{x}}} {x}\sqrt{{y}}\:=\:?\:\:\: \\ $$
Answered by MJS_new last updated on 13/Jul/21
![log_x x(√y) =((ln x(√y))/(ln x))=((ln x +(1/2)ln y)/(ln x))=1+((ln y)/(2ln x))= [y=x^(2/3) ] =1+((ln x^(2/3) )/(2ln x))=1+(((2/3)ln x)/(2ln x))=1+(1/3)=(4/3)](https://www.tinkutara.com/question/Q146460.png)
$$\mathrm{log}_{{x}} \:{x}\sqrt{{y}}\:=\frac{\mathrm{ln}\:{x}\sqrt{{y}}}{\mathrm{ln}\:{x}}=\frac{\mathrm{ln}\:{x}\:+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:{y}}{\mathrm{ln}\:{x}}=\mathrm{1}+\frac{\mathrm{ln}\:{y}}{\mathrm{2ln}\:{x}}= \\ $$$$\:\:\:\:\:\left[{y}={x}^{\mathrm{2}/\mathrm{3}} \right] \\ $$$$=\mathrm{1}+\frac{\mathrm{ln}\:{x}^{\mathrm{2}/\mathrm{3}} }{\mathrm{2ln}\:{x}}=\mathrm{1}+\frac{\frac{\mathrm{2}}{\mathrm{3}}\mathrm{ln}\:{x}}{\mathrm{2ln}\:{x}}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{4}}{\mathrm{3}} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
$${cool}\:{thank}\:{you}\:{Ser} \\ $$