# Logarithmic equation

1. Mar 18, 2014

### Illuvitar

1. The problem statement, all variables and given/known data
$log _3\left(x-5\right) + log_3\left(x+3\right) = 2$

I'm having trouble with applying properties of logarithms to solve equations, I think misunderstanding something fundamental here, I get most of the questions right but there are a handful that I have no idea what I am doing. Please help.
2. Relevant equations

3. The attempt at a solution
I try to condense the logs into one with the product rule so:
$log_3\left(x^2-2x-15\right) = 2$

then I rewrite it in exponential form:
$3^2=x^2-2x-15$

Now I figured that getting rid of the logs would make this quadratic equation manageable but I am coming up nowhere near the right answer (I peaked, its 6).

If I add 15 to both sides I get:
$24=x^2-2x$

but I don't know what to do with the quadratic...I feel like I messed up trying to get rid of the logs but I cant really identify where I screwed up. I tried to rework it a couple times with different tricks but I still am nowhere close. Any help would be appreciated.

2. Mar 18, 2014

### micromass

3. Mar 18, 2014

### Illuvitar

Well I didn't use the quadratic formula but I did try factoring it after subtracting 24 from both sides but I got negative roots and I thought you couldn't have those in a logarithms domain? Sorry... I know this is easy math for alot of you but I am teaching this to myself and am having a bit of trouble. Ill use the quadratic formula thanks.

4. Mar 18, 2014

### IDValour

Just to check, can you please show us the factorisation you did after subtracting 24? There should be one positive and one negative root.

5. Mar 18, 2014

### micromass

Why can't you use the quadratic formula?

An alternative is simply to "complete the square":

For example, take $x^2 +2x - 3=0$. I want to write this as $(x +1)^2 + F=0$ for a certain $F$. Clearly $F=-4$. So the equation becomes $(x+1)^2 -4 = 0$. Thus $(x+1)^2 = 4$. Hence, $x+1 = \pm 4$.

6. Mar 18, 2014

### Illuvitar

Crap. I messed up on the factoring, I should have gotten (x-6)(x+4) but I wrote on my paper (x+6)(x+4). thanks.

And I guess I just didn't think to use the quadratic formula(or completing the square), usually when I use it I end up with complex roots and I think for logs I have to have real non-zero solutions. I'm not sure, I just started learning logs yesterday.

7. Mar 18, 2014

### Illuvitar

Oh and I have one more if you guys dont mind. This is one of the ones where I dont know where to start.
1. The problem statement, all variables and given/known data
$log(x+4)= logx+log4$

2. Relevant equations

3. The attempt at a solution

I used the product rule to condense the right side
$log(x+4)=log(4x)$

but after that I have no idea what to do...

8. Mar 18, 2014

### Staff: Mentor

If the logs of two numbers are equal, then the two numbers are equal.

Also, it's a good idea right at the start, to indicate what numbers are allowed for each of the log expressions. In this problem, you must have x > - 4 for log(x + 4) to be defined, and you must have x > 0 for log(x) to be defined. Together, these mean that x > 0.

9. Mar 18, 2014

### Mentallic

Move the logs to one side Also keep in mind that $a^0=1$ for any positive a.

10. Mar 18, 2014

### Illuvitar

Okay. So the equation becomes
$x+4=4x$
its not a quadratic so I cant factor it into two roots right? Im sorry I still dont know what to do from here...

11. Mar 18, 2014

### Illuvitar

If I move them to one side I get

$log(4x/x+4)=0$ because of the quotient property right? but I don't know how to simplify it further.
And Im not sure I understand how I should use $a^0=1$
its seems like the terms have an implied power of 1.

12. Mar 18, 2014

### Staff: Mentor

Right, it's not a quadratic. Add -x to both sides. This is really a simple equation...

13. Mar 18, 2014

### scurty

$e^0 = ...$

14. Mar 18, 2014

### Illuvitar

Alright sorry...

15. Mar 18, 2014

### Staff: Mentor

You have really made it complicated. Also, I think the fraction is intended to be $\frac{4x}{x + 4}$, so you should have written it as 4x/(x + 4) if that's what you meant. Without the parentheses, knowledgeable people would interpret it as (4x/x) + 4.
Of course. x is the same as x1.

16. Mar 19, 2014

### Mentallic

I woke up too early this morning to start answering questions. Clearly if

$$\log(a)=\log(b)$$ then a=b. Moving everything to one side, using a log property and then using the definition of logs as so:

$$\log(a)=\log(b)$$

$$\log(a)-\log(b)=0$$

$$\log\left(\frac{a}{b}\right)=0$$

$$\frac{a}{b}=1$$

$$a=b$$

Is just a long winded way of showing the same thing.