# Thread: help solving using logarithms..

1. ## help solving using logarithms..

i need to solve this equation for x using base 10 logarithms...

13^(x-3) = 18^(3x)

2. Originally Posted by Manizzle
i need to solve this equation for x using base 10 logarithms...

13^(x-3) = 18^(3x)
so take the log of both sides

$\displaystyle \log 13^{x - 3} = \log 18^{3x}$

now recall the rule: $\displaystyle \log_a (x^n) = n \log_a x$

now proceed

3. yes i know that rule but i don't know how then to solve for x

4. Originally Posted by Manizzle
yes i know that rule but i don't know how then to solve for x
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

5. Originally Posted by Manizzle
i need to solve this equation for x using base 10 logarithms...

13^(x-3) = 18^(3x)
Sorry, Jhevon. But I had already done the grunt work on this so here's the rest of it.

$\displaystyle 13^{x-3}=18^{3x}$

$\displaystyle \log(13^{x-3})=\log (18^{3x})$

$\displaystyle (x-3)\log 13=(3x)\log 18$

$\displaystyle (x-3) \cdot 2.564949357=3x \cdot 2.890371758$

$\displaystyle 2.564949357x-7.694848071=8.671115274x$

$\displaystyle -7.694848071=6.106165917x$

$\displaystyle x=\frac{-7.694848071}{6.106165917}$

$\displaystyle \boxed{x \approx -1.260176709}$

6. Originally Posted by masters
Sorry, Jhevon. But I had already done the grunt work on this so here's the rest of it.

$\displaystyle 13^{x-3}=18^{3x}$

$\displaystyle \log(13^{x-3})=\log (18^{3x})$

$\displaystyle (x-3)\log 13=(3x)\log 18$

$\displaystyle (x-3) \cdot 2.564949357=3x \cdot 2.890371758$

$\displaystyle 2.564949357x-7.694848071=8.671115274x$

$\displaystyle -7.694848071=6.106165917x$

$\displaystyle x=\frac{-7.694848071}{6.106165917}$

$\displaystyle \boxed{x \approx -1.260176709}$
that's fine, Masters. I would have preferred if you left it as log(13) and log(18) though as opposed to finding the decimal approximations.