i need to solve this equation for x using base 10 logarithms...
13^(x-3) = 18^(3x)
ok, so that means you got to the point $\displaystyle (x - 3) \log 13 = 3x \log 18$
now what is the problem? $\displaystyle \log 13$ and $\displaystyle \log 18$ are constants. how would you solve something like $\displaystyle (x - 3) \cdot 5 = 3x \cdot 8$ ?
just treat the logs like constants. they are just numbers. solve for $\displaystyle x$ as you would in any regular algebraic equation
Sorry, Jhevon. But I had already done the grunt work on this so here's the rest of it.
$\displaystyle 13^{x-3}=18^{3x}$
$\displaystyle \log(13^{x-3})=\log (18^{3x})$
$\displaystyle (x-3)\log 13=(3x)\log 18$
$\displaystyle (x-3) \cdot 2.564949357=3x \cdot 2.890371758$
$\displaystyle 2.564949357x-7.694848071=8.671115274x$
$\displaystyle -7.694848071=6.106165917x$
$\displaystyle x=\frac{-7.694848071}{6.106165917}$
$\displaystyle \boxed{x \approx -1.260176709}$