Question

If $y=a^{\left(\frac{1}{1-{\log }_{a}x}\right)}z=a^{\left(\frac{1}{1-{\log }_{a}y}\right)}$ then $x$ is equal to

• A. $1+\log yz$
• B. $\frac{1}{1-{\log }_{a}z}$
• C. $a^{\left(\frac{1}{1-{\log }_{a}z}\right)}$
• D. $\frac{y}{z}$

Answer 1

The correct option is C $a^{\left(\frac{1}{1-{\log }_{a}z}\right)}$

$y=a^{\left(\frac{1}{1-{\log }_{a}x}\right)}$

& $z=a^{\left(\frac{1}{1-{\log }_{a}y}\right)}$

$\Rightarrow z=a^{\left\{\frac{1}{[1-{\log }_{a}a\left(\frac{1}{1-{\log }_{a}x}\right)]}\right\}}$

$\Rightarrow z=a^{\frac{1}{1-\left(\frac{1}{1-{\log }_{a}x}\right){\log }_{a}a}}$

$\Rightarrow z=a^{\left[\frac{1-{\log }_{a}x}{1-{\log }_{a}x-1}\right]}$

$\Rightarrow z=a^{(1-\frac{1}{{\log }_{a}x})}$

Take log on both side

$\Rightarrow {\log }_{a}z=1-\frac{1}{{\log }_{a}x}$

$\Rightarrow {\log }_{a}x=(1-\frac{1}{{\log }_{a}z})\Rightarrow x=\frac{1}{a^{\left(\frac{1-1}{{\log }_{a}z}\right)}}$