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Question Number 146459 by mathdanisur last updated on 13/Jul/21

$$if{x}^2=y^{3}find{log}_{\mathit{x}}x\sqrt{y}=?$$

Answered by MJS_new last updated on 13/Jul/21

$${\log }_{x}x\sqrt{y}=\frac{\ln x\sqrt{y}}{\ln x}=\frac{\ln x+\frac{1}{2}\ln y}{\ln x}=1+\frac{\ln y}{2\ln x}=$$$$[y=x^{2/3}]$$$$=1+\frac{\ln x^{2/3}}{2\ln x}=1+\frac{\frac{2}{3}\ln x}{2\ln x}=1+\frac{1}{3}=\frac{4}{3}$$

Commented by mathdanisur last updated on 13/Jul/21

$$coolthankyouSer$$

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