# Am I wrong ? log question

1. Apr 19, 2013

### lionely

1. The problem statement, all variables and given/known data
2log2 x - log2 (x-3) = 2

The attempt at a solution

So what I did was , expand the brackets

2log x - logx + log3 = 2

logx = 2-log3

logx = log 4 - log 3

log x = log ( 4-3)

x= 4/3?

Is this right?

2. Apr 19, 2013

### Dick

No, it's wrong. log(a-b) is not equal to log(a)-log(b). Try taking 2 to the power of each side of your original equation.

3. Apr 19, 2013

### lionely

Aww got that wrong on my test then ;((

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

removing logs

x^2/(x-3) = 16
16x-48 = x^2

x^2 -16x + 48=0
(x -4 )(x -12)

Last edited: Apr 19, 2013
4. Apr 19, 2013

### Dick

What you get after 'removing logs' would be fine if your original equation were 2log(x)-log(x-3)=4 since 4=log(16). Was it?

5. Apr 19, 2013

### Staff: Mentor

This is wrong, too, right in your first step. I should add that what you wrote is probably not what you meant. This is how the left side would be interpreted:
$$\frac{log x^2}{x - 3}$$

From your subsequent work, I think this is what you meant:
$$log(\frac{x^2}{x - 3})$$

If so, when you write it in plain text (as opposed to LaTeX) add parentheses, like this
log(x2/(x - 3))

Where you headed in your first step uses the idea that if log A = log B, then A = B.
Applying that idea, you get x2/(x - 3) on the left side. What do you get on the right side? Hint: NOT 16.

6. Apr 20, 2013

### lionely

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

(x^2/(x-3)) = 4?

7. Apr 20, 2013

Yes.