Looking for a bit of help on my log rules here, thought I had a relatively idea until trying to solve this:

Prove

$\displaystyle \frac{3^{k - 1} - 1}{2} + 3^{k - 1}$ = $\displaystyle \frac{3^k - 1}{2}$

So choosing to start with the left side and find the right:

$\displaystyle = \frac{3 * 3^{k - 1} - 1}{2}$

$\displaystyle = (ln(3 * 3^{k - 1}) - \ln(2)) - \frac{1}{2}$ - can I just use ln on a portion of the equation like I have here

$\displaystyle = ((ln(3) + (k - 1)\ln(3)) - \ln(2)) - \frac{1}{2}$

$\displaystyle = ((ln(3)(1 + (k - 1))) - \ln(2)) - \frac{1}{2}$

$\displaystyle = ((k\ln(3)) - \ln(2)) - \frac{1}{2}$

$\displaystyle = \frac{3^k - 1}{2}$

Really rusty on this so if anyone can have a quick look and let me know if there are errors in there that'd be great. (also first time using LaTeX)

Thanks