Can Logarithms and Triangle Properties Solve This Complex Mathematical Equation?

[L²Cc]

Homework Statement

Define a, b, and c as the sides of a right triangle where c is the hypotenuse, and a > 1 and c > b+1

show that
log_{c+b} a + log_{c-b} a = 2(log_{c+b} a)(log_{c-b} a)


2. Governing equations

The Attempt at a Solution

Should I assume that a=2 and c=b+2?!

 

[AlephZero]

Should I assume that a=2 and c=b+2?!

What did Mr. Pythagoras say about "the sqaw on the hippopotamus"?

 

[L²Cc]

c²=a²+b²...but then how do I apply that to the given statement. Perhaps if you give me a hint, I will manage to finish off the question.

 

[L²Cc]

?

 

[radou]

I haven't got time to look at it now, but maybe it would be useful, as mentioned above, to use a^2 = c^2 - b^2 = (c - b) (c + b), and the fact that \log_{a}x^n = n\log_{a}x somehow.

 

[AlephZero]

OK, here's the strategy I used to get a handle on this.

First off, it's a totaly 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).