Solve y(a^y-1)=b

Question

Trying to solve this equation for $y$ leads me into horrible thickets of logarithms that also seem unsolvable.

$$y(a^y-1)=b$$

$a$ and $b$ are constants.

Is there a simple solution of the form:

$$y = f( a, b, c... )$$

where c... represents possible other constants.

Answer 1

When question does not give certain values for a and b, it means there can be infinitely many equations of these form. For example:

$y=3, a=5$ results in:

$3(5^3-1)=b$, or $b=372$

And we have:

$3(5^3-1)=372$

Now suppose we have following equation:

$y(5^y-1)=372$

$372=3\times 124$

Then we must have:

$y=3$ which also satisfy second equation $5^y-1=124$

$372=2^2\times 3\times 31 $

So we must check other factors too to find more solution if any.If a, b and y are all integers this is most simple method.