Given that $3^{15a} = 5^{5b} = 15^{3c}$, show that $5ab-bc-3ac=0$ Given that $3^{15a} = 5^{5b} = 15^{3c}$, show that $5ab-bc-3ac=0$
The only thing I can do is:
$3^{5a} = 5^{b} = 15^{\frac{3}{5}c}$
and then i am stuck, I figure that there must be a relationship between 3 and 5? should i utilise $5ab-bc-3ac=0$? 

To get the variables directly take logarithms. (Doesn't matter which base).
$\log 3^{15a} = \log 5^{5b} = \log 15^{3c}$ so 
$15a \log 3 = 5b \log 5 = 3c \log 15$.
So flip a coin and choose which one we should express the others in.  I pick ... $b$.
$a = \frac {\log 5}{3\log 3} b$ and $c  = \frac {5\log 5}{3\log 15} b$
So
$5ab - bc -3ac =$
$\frac {5\log 5}{3\log 3}b^2 - \frac {5\log 5}{3\log 15}b^2 - \frac {15(\log 5)^2}{9\log 3\log 15} b^2 = $
$b^2 (\frac {5\log 5}{3\log 3} - \frac {5\log 5}{3\log 15} - \frac {15(\log 5)^2}{9\log 3\log 15} )$
Which reduces to proving $\frac {5\log 5}{3\log 3} - \frac {5\log 5}{3\log 15} - \frac {15(\log 5)^2}{9\log 3\log 15} = 0$
Which requires comfort with logs.
$\frac {5\log 5}{3\log 3} - \frac {5\log 5}{3\log 15} - \frac {15(\log 5)^2}{9\log 3\log 15}=$
$\frac {5\log 5\log 15 - 5\log 5\log 3}{3\log 3 \log 15}-\frac{5(\log 5)^2}{3\log 3\log 15}=$
$\frac {5\log 5\log 15 - 5\log 5\log 3-5(\log 5)^2}{3\log 3\log 15}=$
$\frac {5[\log 5(\log 3 + \log 5) -\log 5\log 3 -(\log 5)^2] }{3\log 3\log 15}=$
$\frac {5[\log 5\log 3 + (\log 5)^2 - \log 5\log3 - (\log 5)^2]}{3\log 3 \log 15}=$
$\frac {5\cdot 0}{3\log 3 \log 15}=0$
