# Geometric series involving logarithms

1. Jan 28, 2014

### sooyong94

1. The problem statement, all variables and given/known data
A geometric series has first term and common ratio both equal to $a$, where $a>1$
Given that the sum of the first 12 terms is 28 times the sum of the first 6 terms, find the exact value of a.

Hence, evaluate
$log_{3}(\frac{3}{2} a^{2}+ a^{4}+...+ a^{58})$

Giving your answer in the form of $A-log_{3} B$, where $A$ and $B$ are positive integers to be determined.

2. Relevant equations
Geometric series, logarithms

3. The attempt at a solution
For the first part, I have to write them as
$S_{12} =28 S_{6}$. Then I have to apply the formula for geometric sum...
Then I arrived at $a^{13} -28a^{7} +27a=0$
Factoring give me $a(a^{12}-28a^{6}+27)=0$

How should I solve for a in this case?

2. Jan 28, 2014

### Staff: Mentor

If you replace a^6 by b then you'll notice you have a quadratic b^2 - 28b + 27 = 0 and you should be able to solve for b and by extension for a.

3. Jan 28, 2014

### sooyong94

Ok, now I have a=sqrt(3)... But how about the second part? :/

4. Jan 29, 2014

### Saitama

You have the value of a, just plug it in the second part. It's a geometric progress inside the log. :)

5. Jan 29, 2014

### sooyong94

But it doesn't look like one if I plugged in...
I got 9/2 + 9 + 27+...

6. Jan 29, 2014

### ehild

The sum is that of a geometric progression + something.
(1/2 a2+a2+a4+....+a58))

ehild

7. Jan 29, 2014

### sooyong94

I get $30-\log_3 {2}$. Is this correct?

8. Jan 29, 2014

Yes.

ehild