# I need sums in LOG

1. Feb 22, 2006

### dilan

Hi,

I just need a little help in getting some sums. Can anyone of you give me a site where I can find sums in Log so that I can do them and practice a lot.

I mean like sums in this type

Show that log(xy)base16 = 1/2log(X)base4 + 1/2log(Y)base4

Thanks just need some sums of this type to practice my self.

Thanks alot people just give me a few links

Thanks

2. Feb 22, 2006

### arildno

Okay, don't bother with the base change first:
Firstly:
How can you change your left-hand side from a product into a sum?

3. Feb 23, 2006

### dilan

Well I mean

Ok I mean not converting to a sum. I mean to prove that you can convert it to a sum.
I mean to prove only 1 side to get the left hand side. And then show that it could be proved.

I think I expressed in the correct way because I am from a non-english country now learning in the english medium

4. Feb 23, 2006

### arildno

Well, but a fundamental property about any log is that we have log(xy)=log(x)+log(y)

5. Feb 23, 2006

### dilan

Ya ya I know that, but you can convert it to sums like I've shown above isn't it?

6. Feb 23, 2006

### arildno

Let's take it in detail.
We have:
$$\log_{16}(xy)=\log_{16}(x)+\log_{16}(y)$$
by the fundamental property of logs.

Now, we need to relate logs with different bases!
We have, for bases a, b, the identities:
$$x=a^{\log_{a}(x)}=b^{\log_{b}(x)}, a=b^{log_{b}(a)$$
Thus, we get:
$$b^{\log_{b}(x)}=(b^{\log_{b}(a)})^{\log_{a}(x)}=b^{\log_{b}(a)\log_{a}(x)}$$
Since logs are unique, we therefore have:
$$\log_{b}(x)=\log_{b}(a)\log_{a}(x)$$
Now, let b=4, a=16, and get your result.