# Logarithm Investigation Help

1. Nov 1, 2004

### nobb

Hey
I am doing an investigation for logarithms, and I have a question. logx^n = nlogx. Based on previous knowledge of exponents, could someone please explain why this is true? Thanks.

2. Nov 1, 2004

### recon

Let lgx = m. Then, 10m = x.

xr = (10m)r
xr = 10rm
lgxr = rm
lgxr = rlgx

3. Nov 2, 2004

### nobb

I don't really get what you are doing. You are solving it with knowledge of logs. Is it possible to answer the question with knowledge from exponents only? Or could you please tell me how this relates to exponents?

4. Nov 2, 2004

### Galileo

Use the identity:
$$a^x=y \iff x=\log_ay$$
Then with the familiar rules for manipulating exponentials, the rules for logarithms follow automatically.

5. Nov 3, 2004

### recon

OK, I try again.

Let lgx = m. Then,

10m = x. <--- This is taken directly from the definition of logs.

x = 10m

xr = (10m)r <-- follows from the rule that says if a = b, then ar = b r

xr = 10rm <-- follows from the rule that says (ar)m = arm

lgxr = rm <--- from the definition of logs again.

lgxr = rlgx <--- Remember the definition we gave on the first line of this post that states that m = lgx?

6. Nov 3, 2004

### HallsofIvy

There are a number of DIFFERENT ways to prove log(xn)= n log(x)
depending on exactly how you are DEFINING "log(x)". What is your definition of
log(x)??