Question

If $y=a^{\frac{1}{1-{\log }_{a}x}}$ and $z=a^{\frac{1}{1-{\log }_{a}y}}$, then $x=$

• A. $a^{\frac{1}{a-{\log }_{a}z}}$
• B. $a^{\frac{1}{z-{\log }_{a}z}}$
• C. $a^{\frac{1}{1-{\log }_{a}z}}$
• D. $Noneofthese$

Answer 1

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

Given $y=a^{\frac{1}{1-{\log }_{a}x}}$
Taking logarithm with base $a$ on both sides, we get
${\log }_{a}y=\frac{1}{1-{\log }_{a}x}$ .....(1)
Also given ,$z=a^{\frac{1}{1-{\log }_{a}y}}$
Taking logarithm with base $a$ on both sides, we get
${\log }_{a}z=\frac{1}{1-{\log }_{a}y}$
Put the value from eqn (1),
${\log }_{a}z=\frac{1}{1-\frac{1}{1-{\log }_{a}x}}$
$\Rightarrow {\log }_{a}z=\frac{1}{\frac{1-{\log }_{a}x-1}{1-{\log }_{a}x}}$
$\Rightarrow {\log }_{a}z=\frac{1-{\log }_{a}x}{-{\log }_{a}x}$
$\Rightarrow -{\log }_{a}x{\log }_{a}z=1-{\log }_{a}x$
$\Rightarrow {\log }_{a}x-{\log }_{a}x{\log }_{a}z=1$
Taking ${\log }_{a}x$ common in LHS,
$\Rightarrow {\log }_{a}x(1-{\log }_{a}z)=1$
$\Rightarrow {\log }_{a}x=\frac{1}{1-lo{g}_{a}z}$
Converting to exponential form, we get
$x=a^{\frac{1}{1-{\log }_{a}z}}$