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 $5ab-bc-3ac=0$? 
 A: You have
$$3^{15a} = 5^{5b} = 15^{3c} = 3^{3c}5^{3c} \tag{1}\label{eq1A}$$
Taking natural logarithms (although any other logarithm base, e.g., common (i.e., base $10$), will also work) gives
$$15a\ln(3) = 5b\ln(5) = 3c\ln(3) + 3c\ln(5) \tag{2}\label{eq2A}$$
This gives, using the left & right and then the middle & right parts,
$$15a\ln(3) = 3c\ln(3) + 3c\ln(5) \implies (15a - 3c)\ln(3) = 3c\ln(5) \tag{3}\label{eq3A}$$
$$5b\ln(5) = 3c\ln(3) + 3c\ln(5) \implies 3c\ln(3) = (5b - 3c)\ln(5) \tag{4}\label{eq4A}$$
Eliminate the $\ln(3)$ terms by multiplying \eqref{eq3A} by $3c$ and subtracting \eqref{eq4A} multiplied by $15a - 3c$ to get
$$((3c)(3c) - (15a - 3c)(5b - 3c))\ln(5) = 0 \tag{5}\label{eq5A}$$
Since $\ln(5)$ is not $0$, you have
$$\begin{equation}\begin{aligned}
(3c)(3c) - (15a - 3c)(5b - 3c) & = 0 \\
9c^2 - 75ab + 15bc + 45ac - 9c^2 & = 0 \\
-15(5ab - bc - 3ac) & = 0 \\
5ab - bc - 3ac & = 0
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
A: Let $$3^{15a}=5^{5b}=15^{3c}=k$$
What if $k=1$
Else
$3=k^{1/15a}$ etc.
As $15=3\cdot5$
$k^{1/3c}=k^{1/5b}\cdot k^{1/15a}=k^{1/5b+1/15a}$
$\implies\dfrac1{3c}=\dfrac1{5b}+\dfrac1{15a}=\dfrac{3a+b}{15ab}$
A: Take $$3^{15a}=5^{5b}=15^{3c}=K \implies a=\frac{\log k}{15\log 3}, b=\frac{\log K}{5\log 5}, c=\frac{\log K}{3 \log 15} $$
Then $$ F=5 ab-bc-3ac=abc(5/c-1/a-3/b)=anc\left(\frac {15 \log 15}{\log K}-\frac{15 \log 3}{\log K}-\frac{15 \log 5}{\log K}\right)$$ $$=abc\frac{15 \log(15/15)}{\log K}=0.$$
A: 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.
$\frac {5\log 5}{3\log 3} - \frac {5\log 5}{3\log 15} - \frac {15(\log 5)^2}{9\log 3\log 15}=$
$\frac {5\log 5\log 15 - 5\log 5\log 3}{3\log 3 \log 15}-\frac{5(\log 5)^2}{3\log 3\log 15}=$
$\frac {5\log 5\log 15 - 5\log 5\log 3-5(\log 5)^2}{3\log 3\log 15}=$
$\frac {5[\log 5(\log 3 + \log 5) -\log 5\log 3 -(\log 5)^2] }{3\log 3\log 15}=$
$\frac {5[\log 5\log 3 + (\log 5)^2 - \log 5\log3 - (\log 5)^2]}{3\log 3 \log 15}=$
$\frac {5\cdot 0}{3\log 3 \log 15}=0$
