Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Possible Duplicate:
Intuitive use of logarithms

My math teacher "taught" us about logarithms today, but he didn't give any useful information. He just that one is supposed to "add" them to create a quadratic equation. He then gave us this example;

$$ \log_4 (x + 4) + \log_4 (x - 4) $$

He then told us to solve it. This is how he did it.

$$\log_4 (x + 4) + \log_4 (x - 4)$$

$$(x + 4)(x - 4) = 0$$

$$x = -4, 4$$

This does not make a bit of sense to me. What are logarithms for? What do they do? And, most importantly, how would I actually solve this equation?

share|cite|improve this question

marked as duplicate by Andrés E. Caicedo, t.b., J. M., Sivaram, Qiaochu Yuan May 10 '11 at 5:42

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I just want to point out, that it is sloppy to say "solve" since "$\log_4$(x + 4) + $\log_4$(x - 4)" is technically $not$ an equation. – Tyler May 10 '11 at 5:02
Doesn't make a lot of sense to me either. For one thing, an equation needs to have an $=$ sign, and your first line doesn't. For general information on logarithms, please see (You will want to skip initially the parts of this article that are clearly too advanced.) – André Nicolas May 10 '11 at 5:02
You'll also want to note that $x = -4$ is an extraneous solution since you can't take the logarithm of a negative number in the original equation (in the Reals) – Tyler May 10 '11 at 5:08
I know that all of you say it is not right at all, bit this is exactly what he wrote. That is why I am confused. Thanks for all of the help and links! – Justin May 10 '11 at 5:08
up vote 5 down vote accepted

If you want to solve $$\log_4{(x-4)} + \log_4{(x+4)} = 0$$ you combine that to $$\log_4{(x-4)(x+4)} = 0$$ then exponentiate each side by taking $$4^{\log_4{(x-4)(x+4)}} = 4^0$$ which gives $$ (x-4)(x+4) = 1 $$ which you can solve using whichever algebraic method you prefer.

You should get $x = -\sqrt{17}, +\sqrt{17}$, but as I noted in my above comment, the negative solution is extraneous.

share|cite|improve this answer
These are all "Logarithmic Identities" which you can read about on the wikipedia page. – Tyler May 10 '11 at 5:18
Thanks. I think I am beginning to understand this. – Justin May 10 '11 at 5:20
You should use braces in the first equation – Listing May 10 '11 at 5:20
@user3123: thanks, I noticed that after I posted and have fixed it, after frustratingly losing my internet connection in the middle of the edit :p – Tyler May 10 '11 at 5:27

Not the answer you're looking for? Browse other questions tagged or ask your own question.