OK, here's the strategy I used to get a handle on this.
First off, it's a totally weird looking equation. I can't think what practical use it would be. The things in there are "a", "c+b" and "c-b" It's also got logs to two different bases in the same equation.
I tried putting in the numbers for some rightangled triangles like 3,4,5 and 5,12,13. That didn't help much.
What else are you given? some weird stuff, a > 1 and c > b+1 or c-b > 1. Hm... maybe that's just saying all the "interesting" things are positive numbers so the logs are well defined. I don't know what else to do with it yet, so forget about it till later...
What else do we know? Well, a^2 + b^2 = c^2 is the only equation you have. So try and work forwards from a^2 + b^2 = c^2, and backwards from what you are trying to prove, and see if you can meet up in the middle.
From a^2 + b^2 = c^2 we want some stuff with c+b, c-b, and some logs in it. OK...
a^2 = c^2 - b^2 = (c+b)(c-b)
2 ln a = ln (c+b) + ln(c-b)
... that looks promising but all the logs are to the same base (e). So what happens if you work backwards from the answer and transform all the logs into natural logs? The standard formula is log(base p)q = ln(q)/ln(p).