If y = a^(1/{1-log_a x}) Z = a^(1/{1-log_a y}) prove that x = a^(1/{1-log_a z}) ?

Aug 27, 2015

Have a look:

Explanation:

Let us take ${\log}_{a}$ of the first to get:
${\log}_{a} y = \frac{1}{1 - {\log}_{a} x}$
And also of the second:
${\log}_{a} z = \frac{1}{1 - {\log}_{a} y}$
Substitute the first into the second:
${\log}_{a} z = \frac{1}{1 - \frac{1}{1 - {\log}_{a} x}}$
Rearrange:
${\log}_{a} z = \frac{1}{\frac{\cancel{1} - {\log}_{a} x \cancel{- 1}}{1 - {\log}_{a} x}}$ change sign on the right and rearrange again:
${\log}_{a} z {\log}_{a} x = {\log}_{a} x - 1$
${\log}_{a} z {\log}_{a} x - {\log}_{a} x = - 1$
Collect ${\log}_{a} x$:
${\log}_{a} x \left[{\log}_{a} z - 1\right] = - 1$
Change sign and isolate ${\log}_{a} x$:
${\log}_{a} x = \frac{1}{1 - {\log}_{a} z}$
Take the power of $a$ on both sides:
$x = {a}^{\frac{1}{1 - {\log}_{a} z}}$

Aug 27, 2015

Require theory:
${\log}_{c} {b}^{n} = n {\log}_{c} b$
${\log}_{c} c = 1$

From question:
$y = {a}^{\frac{1}{1 - {\log}_{a} x}}$ ... (Eq 1)
$z = {a}^{\frac{1}{1 - {\log}_{a} y}}$ ...(Eq 2)

Take log on both sides of equation 2:
${\log}_{a} z = \frac{1}{1 - {\log}_{a} y}$
$1 - {\log}_{a} y = \frac{1}{\log} _ a z$

Substitute expression for $y$:
$1 - \frac{1}{1 - {\log}_{a} x} = \frac{1}{\log} _ a z$
${\log}_{a} \frac{z}{{\log}_{a} z - 1} = 1 - {\log}_{a} x$
${\log}_{a} x = 1 - {\log}_{a} \frac{z}{{\log}_{a} z - 1} = - \frac{1}{{\log}_{a} z - 1}$
$x = {a}^{\frac{1}{1 - {\log}_{a} z}}$