1. Log rule confusion

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

2. Re: Log rule confusion

Skip the logarithms...

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

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

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

3. Re: Log rule confusion

That's much easier, can you explain the basis of going from: $\displaystyle 3(3^{k-1})$ to $\displaystyle 3^{k}$ ?

4. Re: Log rule confusion

Hint: $\displaystyle a^b a^c = a^{b+c}$.

5. Re: Log rule confusion

I've been doing too much math today, I was looking at that as $\displaystyle ({3^{k-1}})^{3}$

Thanks for that!