1. ## Logs

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?

2. Originally Posted by RFB4191
So I have this equation:

(1/27)^(x-1) = 3 times cubed root of nine

I need to solve 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

3. Originally Posted by RFB4191
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?
$\displaystyle \left ( \frac{1}{27} \right ) ^{x - 1} = 3 \sqrt[3]{9}$

$\displaystyle 27^{-(x - 1)} = 3 \cdot 3^{2/3}$

$\displaystyle 3^{-3(x - 1)} = 3^{5/3}$

Can you take it from here?

-Dan

4. Originally Posted by topsquark
$\displaystyle \left ( \frac{1}{27} \right ) ^{x - 1} = 3 \sqrt[3]{9}$

$\displaystyle 27^{-(x - 1)} = 3 \cdot 3^{2/3}$

$\displaystyle 3^{-3(x - 1)} = 3^{5/3}$

Can you take it from here?

-Dan

I got it... thanks a lot