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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?

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Justin
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    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 05:02
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    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 http://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 05:02
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    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 05:08
  • 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 05:08

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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.

Tyler
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