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?

Printable View

- May 15th 2008, 06:51 PMRFB4191Logs
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? - May 15th 2008, 06:59 PMMathstud28
$\displaystyle \bigg(\frac{1}{27}\bigg)^{x-1}=\frac{1}{27^{x-1}}=\frac{1}{3^{3x-3}}=3^{-(3x-3)}$

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 - May 15th 2008, 07:00 PMtopsquark
- May 15th 2008, 07:12 PMRFB4191