So I have this equation:
(1/27)^(x-1) = 3 times cubed root of nine
I need to solve for x this using logs...
Any thoughts?
$\displaystyle \bigg(\frac{1}{27}\bigg)^{x-1}=\frac{1}{27^{x-1}}=\frac{1}{3^{3x-3}}=3^{-(3x-3)}$
also note that $\displaystyle 3\cdot\sqrt[3]{9}=3\cdot\sqrt[3]{3^2}=3\cdot{3^{\frac{2}{3}}}=3^{\frac{5}{3}}$
So now we have $\displaystyle 3^{-(3x-3)}=3^{\frac{5}{3}}$
Taking the $\displaystyle \log_3$ of both sides we get
$\displaystyle -(3x-3)=\frac{5}{3}$
I think you can take it from there