Question

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

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

Answer 1

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

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