# Numerator log help

1. Apr 29, 2009

### tweety1234

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

$$log_4(5-x)-log_4(3-x) = 2$$

3. The attempt at a solution

$$log_4(5-x)-log_4(3-x) = 2$$
$$log_4\frac{3-x}{5-x} = 2$$

where do I go from here?

2. Apr 29, 2009

### Cyosis

Re: logarithms

What do you know about $$a^\log_a x$$?

3. Apr 29, 2009

### tweety1234

Re: logarithms

Dont know, not come across that expression before. Could you please explain.

Thanks.

Are you using the power rule ?

4. Apr 29, 2009

### Chewy0087

Re: logarithms

It's actually much simpler than that cyosis.

You're on the right track but what do you know about logs?

Consider this;

log 2 (4) = 2 Am I right? Isn't this just another way of saying, 2^2 = 4?

Ok, let's apply this logic to your question.

log 4 (3-x / 5-x ) = 2

Now do the same

Last edited: Apr 29, 2009
5. Apr 29, 2009

### HallsofIvy

Staff Emeritus
Re: logarithms

$log_a(b)= c$ is equivalent to $b= a^c$. Here, a= 4, b= (3-x)/(5- x), and c= 2.

6. Apr 29, 2009

### tweety1234

Re: logarithms

Oh right I get it now,
thanks

$$\frac{3-x}{5-x} = 4^2$$

7. May 1, 2009

### DorianG

Re: logarithms

Shouldn't the 5-x term be in the numerator and the 3-x term in the denominator?
I'm not 100% certain, but I thought log b - log a = log (b/a), not log (a/b)?

8. May 1, 2009

### Chewy0087

Re: logarithms

Yeah you're right