logs!

**rtwilton** · November 24th 2008, 07:54 AM

Can anyone give me a hand with this one?

loga(x) + loga(x - 2) = loga(x + 4)

**masters** · November 24th 2008, 08:41 AM

Originally Posted by rtwilton

Can anyone give me a hand with this one?

loga(x) + loga(x - 2) = loga(x + 4)

Hello rt,

Remember $\log_bm + \log_bn = \log_bmn$

$\log_ax+\log_a(x-2)=\log_a(x+4)$

$\log_ax(x-2)=\log_a(x+4)$

$x(x-2)=x+4$

Can you finish from here?

**rtwilton** · November 24th 2008, 08:54 AM

I think so. I distributed it out, got everything on one side and got:

x^2 -3x - 4 = 0

so x = -1 and x = 4

but x cannot be a negative number right? so x=4 is the answer?

**masters** · November 24th 2008, 09:17 AM

Originally Posted by rtwilton

I think so. I distributed it out, got everything on one side and got:

x^2 -3x - 4 = 0

so x = -1 and x = 4

but x cannot be a negative number right? so x=4 is the answer?

You can take the log of a negative number,
but your answer is no longer "real", but complex.

A logarithm is the inverse of a power.

If $y=10^x$ , then $x=\log(y)$ .

$10^x$ is always positive. As a result, you can only take the log of a positive number.

However, 10 raised to an imaginary number can be negative, so the log of a negative number is imaginary.

So long as you use restrict your domain to real numbers, the log of a negative number is undefined.

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