What Is the Solution to This Logarithmic Problem?

[MajikWaffle]

http://img228.imageshack.us/img228/1799/untitled8av.png


Kinda Tricky

 

[Tide]

Not too tricky - it's just the Golden Ratio.

 

[JasonRox]

http://img228.imageshack.us/img228/1799/untitled8av.png
Kinda Tricky

A nice little question.

I'll give a hint.

9*16=144

12*12=144

 

[MajikWaffle]

Yar, still can't get it. Answer anybody

 

[Tide]

The answer is in #2 above.

 

[Avodyne]

The problem is to find $a/b$ given
$$\log_9 a = \log_{12}b = \log_{16}(a+b)$$

Let $$x = \log_9 a = \log_{12}b = \log_{16}(a+b)$$. Then

$$a=9^x,~~~b=12^x,~~~a+b=16^x$$

Now, compare $a(a+b)$ with $b^2$.

 

[berkeman]

Yar, still can't get it. Answer anybody

We do not give out answers here at the PF. Show some effort, or your thread will be deleted.