Question

If $a^{x^y}=(a^x{)}^y$, then $x^{y-1}.y^{x-1}=$____

• A. $y^x$
• B. $x^y$
• C. xy
• D. $\frac{y}{x}$

Answer 1

The correct option is A $y^x$

Given $a^{x^y}=(a^x{)}^y$

$\Rightarrow a^{x^y}=a^{xy}$

Apply logarithm on both sides with base $a$

$\Rightarrow {\log }_a{a}^{x^y}={\log }_a{a}^{xy}$

$\Rightarrow x^y\left({\log }_{a}a\right)=xy\left({\log }_{a}a\right)$ $\dots [\because \log t^m=m\log t]$

$\Rightarrow x^y\left(1\right)=xy(1$ $\dots [\because {\log }_{a}a=1]$

$\Rightarrow x^y=xy$

$\Rightarrow \frac{x^y}{x}=y$

$\Rightarrow x^{y-1}=y$ $\dots \left(i\right)$

Required $x^{y-1}.y^{x-1}=y.y^{x-1}$ $[from$ $\left(i\right)]$

$x^{y-1}.y^{x-1}=y^{1+x-1}$

$x^{y-1}.y^{x-1}=y^x$