Question

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

Answer 1

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

$:z=a^{\frac{1}{1-{\log }_{a}y}}$.......$\left(2\right)$

Take $\log$ w.r.t to base $a$ on both sides in $\left(1\right),\left(2\right)$

${\log }_{a}y=\frac{1}{1-{\log }_{a}x}$.....$\left(3\right)$

${\log }_{a}z=\frac{1}{1-{\log }_{a}y}$.....$\left(4\right)$

Let $u={\log }_{a}x,v={\log }_{a}y,w={\log }_{a}z$

So $\left(3\right),\left(4\right)$ will now look like

$v=\frac{1}{1-u}$

$w=\frac{1}{1-v}$

Put value of $v$ in $w$,

$w=\frac{1}{1-\frac{1}{1-u}}\Rightarrow w=\frac{u-1}{u}\Rightarrow w=1-\frac{1}{u}$

$u=\frac{1}{1-w}$

Re substitute $u={\log }_{a}x,w={\log }_{a}z$

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

Hence proved.