Question

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

Answer 1

Have a look:

Explanation:

Let us take `log_a` of the first to get:
`log_ay=1/(1-log_ax)`
And also of the second:
`log_az=1/(1-log_ay)`
Substitute the first into the second:
`log_az=1/(1-1/(1-log_ax))`
Rearrange:
`log_az=1/((cancel(1)-log_axcancel(-1))/(1-log_ax))` change sign on the right and rearrange again:
`log_azlog_ax=log_ax-1`
`log_azlog_ax-log_ax=-1`
Collect `log_ax`:
`log_ax[log_az-1]=-1`
Change sign and isolate `log_ax`:
`log_ax=1/(1-log_az)`
Take the power of `a` on both sides:
`x=a^(1/(1-log_az))`

Answer 2

Require theory:
`log_c b^n = n log_c b`
`log_c c = 1`

From question:
`y = a^(1/(1-log_a x))` ... (Eq 1)
`z = a^(1/(1-log_a y))` ...(Eq 2)

Take log on both sides of equation 2:
`log_a z = 1/(1-log_a y)`
`1 - log_a y = 1/log_a z`

Substitute expression for `y`:
`1 - 1/(1-log_a x) = 1/log_a z`
`log_a z/(log_a z - 1) = 1-log_a x`
`log_a x = 1 - log_a z/(log_a z - 1) = -1/(log_a z - 1)`
`x = a^(1/(1-log_a z))`