# Logs

• Oct 15th 2012, 06:21 AM
RD33
Logs
Solve for x: (9/4)log4x=10

this is what I got:
log4 (x^(9/4))=10
4^10 = x^(9/4)
(4^10)^(4/9) = (x^(9/4))^(4/9)
4^(4/10) = x
4^(2/5) =x

am I on the right track?
• Oct 15th 2012, 06:37 AM
topsquark
Re: Logs
Quote:

Originally Posted by RD33
Solve for x: (9/4)log4x=10

this is what I got:
log4 (x^(9/4))=10
4^10 = x^(9/4)
(4^10)^(4/9) = (x^(9/4))^(4/9)
4^(4/10) = x
4^(2/5) =x

am I on the right track?

First of all I presume this is log base 4? I'd recommend that you start this way:
$\displaystyle \frac{9}{4}log_4x = 10$

$\displaystyle log_4x = \frac{4}{9} \cdot 10$

$\displaystyle log_4x = \frac{40}{9}$

Can you work out the rest?

-Dan
• Oct 15th 2012, 07:08 AM
RD33
Re: Logs
I think so. I would have to put it into exponential from right? So I would end up with 4^(40/9) = x ?
• Oct 15th 2012, 07:18 AM
Soroban
Re: Logs
Hello, RD33!

Quote:

$\displaystyle \text{Solve for }x\!:\;\;\tfrac{9}{4}\log_4x\:=\:10$

This is what I got:

. . $\displaystyle \log_4\left(x^{\frac{9}{4}}\right) \:=\:10$

. . $\displaystyle 4^{10} \:=\: x^{\frac{9}{4}}$

. . $\displaystyle \left(4^{10}\right)^{\frac{4}{9}} \:=\: \left(x^{\frac{9}{4}}\right)^{\frac{4}{9}}$

. . $\displaystyle 4^{\color{red}{\frac{4}{10}}} \:=\: x \;\;\color{red}{??}$

. . $\displaystyle 4^{\frac{2}{5}}\:=\: x$

Am I on the right track?
Just one error in arithmetic.

I solved it like this:

. . $\displaystyle \tfrac{9}{4}\log_4x \:=\:10$

. . . $\displaystyle \log_4x \:=\:\tfrac{40}{9}$

. . . . . . $\displaystyle x \:=\:4^{\frac{40}{9}}$
• Oct 15th 2012, 07:29 AM
RD33
Re: Logs
Ok thank you