# Another log problem

1. Jul 27, 2007

### Cmunro

Hi, I've been trying to work out:

loga(2+x) = 2 + logax

This is as far as I can get:

loga(2+x) - logax =2
so: loga((2+x)/x) =2
loga((2+x)/x) =log100 (I'm not even sure where I could go with this one though).

Relevant equations include:

log x +log y = log xy
log x - log y = log x/y
logax=(logcx)/logca

I'd really appreciate any hints you can give me. Thanks.

2. Jul 27, 2007

### derekjn

It looks like you represented the 2 on the right side as a common log of 100, but the left side is log base a, so that isn't very useful in this case. Since log base a ((2+x)/x) = 2, try thinking more in terms of, "a raised to the second power is equal to ((2+x)/x)."

3. Jul 27, 2007

### Cmunro

I tried to follow this trail of thought, but it didn't seem to take me anywhere:

a^2 =(2+x)/x

xa^2=2+x or alternatively a=root((2+x)/x)

..but I can't see how any of this will get rid of the a, which is ideally what I'd like to do to solve for x.

4. Jul 27, 2007

### Dick

You don't want to get rid of the a. You want to solve for x in terms of a. Got that? Solve for x, not a.

5. Jul 27, 2007

### Cmunro

ohhh I see! I've been quite thick really, of course I don't want to get rid of the a.

Ok so: a$$^{2}$$=2+x/x
xa$$^{2}$$-x=2
x(a$$^{2}$$-1) =2
x= 2/(a$$^{2}$$-1)

Is this right? If yes, can it be simpler?

6. Jul 27, 2007

### Dick

It looks plenty simple to me.

7. Jul 27, 2007

### Cmunro

Thanks for your help both of you :)