# Solve this logarithm

1. Mar 9, 2015

### dylanjames

• Moved from a technical math section, so missing the template
I need to solve log3(2x+3)-log3(x-2)=3 where 3 is the base.

This is my attempt at a solution..
Log3(2x+3)-log3(x-2)=2
Log3(x-3/2)=2
Log3(x) - Log3(-3/2) = 2
Log3(x)-0.369=2
log3(x)=2.369

3^2.369=13.4
x=13.4

I plugged that into the original equation and I know it is not correct. Can anyone try to steer me in the right direction?
Been a while on this log business..

2. Mar 9, 2015

### Staff: Mentor

What did you do between those lines?
The following step is wrong as well.

Last edited: Mar 9, 2015
3. Mar 9, 2015

### Quantum Defect

The difference between two logs is the same thing as the log of the quotient.

log A - log B = log(A/B)

Also, in the top you have the right hand side equal to "3," but in your work, you have this equal to "2" ...

4. Mar 9, 2015

### dylanjames

Ok, let me try that again.

Equation is log3(2x+3)-log3(x-2)=2
log3((2x/x)+(3/-2))=2
log3(2x-1.5)=2
log3(2x-1.5)=2
Into exponential form, 3^2=2x-1.5
X=3.75

But I solved that out and it doesnt seem to work either.

5. Mar 9, 2015

### Quantum Defect

I get log3[(2x+3)/(x-2)] for the first bit of the l.h.s.

6. Mar 9, 2015

### dylanjames

Which would give you log3(2x-3/2), correct?

7. Mar 9, 2015

### Staff: Mentor

No it would not.

Okay, here is an important subproblem: simplify (2x+3)/(x-2)

8. Mar 9, 2015

### Staff: Mentor

The first line above can be rewritten as
$\log_3 \frac{2x + 3}{x - 2} = 2$
So far, so good.
In the second line it looks like you did the division this way:
$\frac{2x + 3}{x - 2} = \frac{2x}{x} - \frac{3}{2} = 2$
If that's what you did, it is incorrect - that's not how you do division. For example, $\frac{4 + 2}{2 + 1} \neq \frac 4 2 + \frac 2 1 = 2 + 2 = 4$. The correct answer is 6/3 = 2.