Question

If $x=\frac{y}{(1+x{)}^p}$, then $p$ is equal to

• A. $\frac{{\log }_e\left(\frac{y}{x}\right)}{{\log }_e(1+a)}$
• B. $\log \left\{\frac{y}{x(1+a)}\right\}$
• C. $\log \left\{\frac{y-x}{1+a}\right\}$
• D. $\frac{\log y}{\log \left\{x\right(1+a\left)\right\}}$

Answer 1

The correct option is A $\frac{{\log }_e\left(\frac{y}{x}\right)}{{\log }_e(1+a)}$

Since $x=\frac{y}{(1+a{)}^p}$

$\therefore (1+a{)}^p=\frac{y}{x}$

or $p{\log }_e(1+a)={\log }_e\frac{y}{x}$

or $p=\frac{lo{g}_e\left(\frac{y}{x}\right)}{lo{g}_e(1+a)}$