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?
Hello, RD33!
$\displaystyle \text{Solve for }x\!:\;\;\tfrac{9}{4}\log_4x\:=\:10$
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 $
. . . $\displaystyle \log_4x \:=\:\tfrac{40}{9}$
. . . . . . $\displaystyle x \:=\:4^{\frac{40}{9}}$