# What are logarithms, and what do they do? [duplicate]

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?

## marked as duplicate by Andrés E. Caicedo, t.b., J. M. is a poor mathematician, user3302, Qiaochu YuanMay 10 '11 at 5:42

• 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 en.wikipedia.org/wiki/Logarithm. (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

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.