# Logarithms help

1. Mar 18, 2009

### tweety1234

1. The problem statement, all variables and given/known data

I cant seem to get my head round this problem, I know how to use the 'change the base rule'

$$log_2 x + log_4x = 2$$

$$\frac{logx}{log2} + \frac{logx}{log4} = 2$$

why is this not correct ??

2. Mar 18, 2009

### mgb_phys

It looks correct - what's the problem
(of course everytime I stray into maths I make an dumb mistake!)

3. Mar 18, 2009

### tweety1234

well the answer book gives the solution like this,

$$log_2x + \frac{log_2 x}{log_2 4} = 2$$ dont understand why its base 2 here?

$$log_2 x + \frac{log_2 x}{2} =2$$

$$\frac{3}{2}log_2 x = 2$$ dont understand were 3/2 came from ?

$$log_2 x = \frac{4}{3}$$

$$x = 2^{\frac{4}{3} }$$

4. Mar 18, 2009

### mgb_phys

They are just collapsing it all into the same base log.

$$log_2 x + \frac{log_2 x}{2} =2$$ is just

$$1.5 * log_2 x =2$$ which is

$$\frac{3}{2} * log_2 x =2$$

It's the same answer as you get - but you can do it this way without needing to calculate log() of anything

5. Mar 18, 2009

### tweety1234

so is this method still correct? can I just add both logs up ?

6. Mar 18, 2009

### mgb_phys

Except you are trying to find X so first you have to multiply out log(2) and log(4)

7. Mar 18, 2009

### tweety1234

sorry, I still don't get where 1.5 comes from ? I thought you just multiply both sides by two to get rid of the fraction ?

$$2log_2 x + log_2 x = 4$$

8. Mar 18, 2009

### mgb_phys

$$log_2 x + \frac{log_2 x}{2} = 2$$

Which if you ignore the logs for now is just; a + a/2 = 2

a(1+1/2) = 2

1.5a = 2

3/2 a =2

9. Mar 18, 2009

### tweety1234

oh I get it now, thanks a lot for your help!!