Question

If $y={\log }_e{\left(\frac{x}{a+bx}\right)}^x$, then x³ y₂ =

(a) (xy₁ − y)²
(b) (1 + y)²
(c) ${\left(\frac{y-x{y}_1}{y_1}\right)}^2$
(d) none of these

Answer 1

(a) (xy₁ − y)²

Here,
$$\begin{gathered} y={\log }_e{\left(\frac{x}{a+bx}\right)}^x \\ \Rightarrow y=x{\log }_e\left(\frac{x}{a+bx}\right) \\ \Rightarrow y_1={\log }_e\left(\frac{x}{a+bx}\right)+x\times \frac{a+bx}{x}\left(\frac{1}{a+bx}-\frac{bx}{{\left(a+bx\right)}^2}\right) \\ \Rightarrow y_1={\log }_e\left(\frac{x}{a+bx}\right)+\left(\frac{a}{a+bx}\right)...\left(1\right) \\ \Rightarrow y_1=\frac{y}{x}+\left(\frac{a}{a+bx}\right)\left[\because y=x{\log }_e\left(\frac{x}{a+bx}\right)\right] \\ \Rightarrow \frac{x{y}_1-y}{x}=\frac{a}{a+bx}...\left(2\right) \\ \mathrm{Differentiating}\left(1\right)\mathrm{we}\mathrm{get}, \\ {y}_2=\frac{a+bx}{x}\left(\frac{a+bx-bx}{{\left(a+bx\right)}^2}\right)-\frac{ba}{{\left(a+bx\right)}^2} \\ \Rightarrow y_2=\frac{a}{x\left(a+bx\right)}-\frac{ba}{{\left(a+bx\right)}^2} \\ \Rightarrow y_2=\frac{a\left(a+bx\right)-abx}{x{\left(a+bx\right)}^2} \\ \Rightarrow y_2=\frac{a^2}{x{\left(a+bx\right)}^2} \\ \Rightarrow y_2=\frac{{\left(x{y}_1-y\right)}^2}{x^3}\left[\mathrm{Using}\left(2\right)\right] \\ \Rightarrow x^3{y}_2={\left(x{y}_1-y\right)}^2 \end{gathered}$$