# help solving using logarithms..

• October 9th 2008, 11:42 AM
Manizzle
help solving using logarithms..
i need to solve this equation for x using base 10 logarithms...

13^(x-3) = 18^(3x)
• October 9th 2008, 12:07 PM
Jhevon
Quote:

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

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

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

now proceed
• October 9th 2008, 12:09 PM
Manizzle
yes i know that rule but i don't know how then to solve for x
• October 9th 2008, 12:17 PM
Jhevon
Quote:

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

now what is the problem? $\log 13$ and $\log 18$ are constants. how would you solve something like $(x - 3) \cdot 5 = 3x \cdot 8$ ?

just treat the logs like constants. they are just numbers. solve for $x$ as you would in any regular algebraic equation
• October 9th 2008, 12:21 PM
masters
Quote:

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.

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

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

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

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

$2.564949357x-7.694848071=8.671115274x$

$-7.694848071=6.106165917x$

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

$\boxed{x \approx -1.260176709}$
• October 9th 2008, 12:26 PM
Jhevon
Quote:

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

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

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

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

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

$2.564949357x-7.694848071=8.671115274x$

$-7.694848071=6.106165917x$

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

$\boxed{x \approx -1.260176709}$

that's fine, Masters. I would have preferred if you left it as log(13) and log(18) though :p as opposed to finding the decimal approximations.