1. ## Log solve confusion

Solving for x:

log4(x+3)+log4(x-3)=2

I'm doing something incorrect, I have gotten answers working through this as x=5 and x=5,-5

I'm trying to understand why, and if in the end I would need to enter as a correct solution x=5 OR x=5,-5

2. ## Re: Log solve confusion

**EDIT For the record, the OP originally posted this: $\displaystyle \log_4(x+3)=\log_4(x-3)=2$, which is what this response is based on **
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Do you actually mean this: one problem is $\displaystyle \log_4(x+3) = 2$, and a second problem is $\displaystyle \log_4(x-3) = 2$? Because there is no way that $\displaystyle \log_4(x-3) = \log_4 (x+3)$Assuming this is what you mean, then to solve for x remember the definition of a log: if $\displaystyle \log_a b = c$ then $\displaystyle a^c = b$. So for the first problem you have $\displaystyle 4^2 = x+3$ so x = 13. How do you get 5?

3. ## Re: Log solve confusion

Solving for x:l
og4(x+3)=log4(x-3)=2
I'm doing something incorrect, I have gotten answers working through this as x=5 and x=5,-5
I'm trying to understand why, and if in the end I would need to enter as a correct solution x=5 OR x=5,-5

I think that you mean $\displaystyle \log_4(x+3)+\log_4(x-3)=2$.

That gives $\displaystyle (x+3)(x-3)=4^2$ or $\displaystyle x^2-9=16$.

So $\displaystyle x=\pm 5~.$

4. ## Re: Log solve confusion

Originally Posted by Plato
I think that you mean $\displaystyle \log_4(x+3)+\log_4(x-3)=2$.
Good detective work!

Originally Posted by Plato
So $\displaystyle x=\pm 5~.$
But if x = -5 then that means you have $\displaystyle \log_4 (-2) + \log_4 (-8) = 2$, which isn't true. The answer is limited to x = +5.

5. ## Re: Log solve confusion

ok, so my work that I did for:

Solving for x:

log4(x+3)+log4(x-3)=2

would be 5,-5 or just 5?

Sorry the response confused me?

From reading, it seems it SHOULD be just "5"?

6. ## Re: Log solve confusion

gbrad88 - since you asked this in a pre-university algebra forum, can I assume that you have not covered complex numbers in your math classes? If that's so, then the logarithm of negative numbers is not something to consider, and for now you should limit your answers such that (x+3) and (x-3) are postive numbers only.

7. ## Re: Log solve confusion

My apologies. I just found this site and posted in the wrong place. Perhaps a mod can use it. I'm actually in a college algebra class. I fell a little behind due to taking multiple classes and a full work schedule and Im playing catch up now. Its not easy being an older guy that went back to college.

So, looking at x = -5 and plugging it in then that means you something that isn't true. I think I understand it better now that I have written it out again. Correct me if Im wrong. THANK YOU for clearing this up.