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?

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- Oct 15th 2012, 06:21 AMRD33Logs
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 AMtopsquarkRe: Logs
- Oct 15th 2012, 07:08 AMRD33Re: 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 AMSorobanRe: 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?

Your reasoning is excellent.

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 AMRD33Re: Logs
Ok thank you