# Quick Log Question!

1. Apr 6, 2009

### juliany

1. The problem statement, all variables and given/known data
Solve: log3^(2x-9)-2log3^x=-2

2. Relevant equations
None

3. The attempt at a solution
I am confused with one part of this equation.
With the 2log3^x, can you move the x to the front to make it 2x log3?

2. Apr 6, 2009

### dx

Yes, that's valid.

3. Apr 6, 2009

### juliany

So that would make it 2x-9log3-2xlog3=-2
Therefore: 2x-9-2x=-2?

4. Apr 6, 2009

### dx

where did the log 3 go in your last step?

5. Apr 6, 2009

### juliany

Woops, was thinking the same base rule.

Can you go: log3^(2x-9)-log3^x^2=-2
Therefore: Log3 (2x-9)/x^2=-2?

6. Apr 6, 2009

### dx

$$2 \log 3^{x}$$ is $$\log 3^{2x}$$. Just to make sure, was this the original equation?

$$\log3^{2x-9}-2\log 3^x=-2$$​

Because it cannot be solved for x.

Last edited: Apr 6, 2009
7. Apr 6, 2009

### juliany

Yes it was, how did you figure out that you can't solve for x?

8. Apr 6, 2009

### dx

Using the rule $\log a^{b} = b \log a$, the equation becomes $$(2x - 9 - 2x) \log 3 = - 2$$, and the 2x and -2x cancel out. Then you get the equation $$\log 3 = 2/9$$. This is true for an appropriate base of the logarithm, which you can find using a log table.

9. Apr 6, 2009

### juliany

Ok, thanks alot.