Logarithm Equation

**Lil** · October 24th 2010, 09:09 AM

I have this one. It's looks easy to solve. I'm trying to solve and nothing. I can't get an answer.
$\frac{1}{2}lg(x^2 +x-5)=lg5x+lg\frac{1}{5x}$
...
$\frac{1}{2}lg(x^2 +x-5)=lg1$ Can I divide /:1/2

**Jhevon** · October 24th 2010, 09:14 AM

Originally Posted by Lil

I have this one. It's looks easy to solve. I'm trying to solve and nothing. I can't get an answer.
$\frac{1}{2}log(x^2 +x-5)=lg5x+lg\frac{1}{5x}$

You need two rules and one fact here:

First, the fact that $\displaystyle \log x$ is a one-to-one function. So,

$\displaystyle \log A = \log B \implies A = B$

Secondly, the rules:

$\displaystyle n \log X = \log (X^n)$

And $\displaystyle \log X + \log Y = \log (XY)$

What can you do with those? I'd apply the rules first before the fact

**Lil** · October 24th 2010, 09:17 AM

I know this rules... I think, I something do wrong.
$lg(x^2 +x-5)^{\frac{1}{2}}=lg1$

From here I stuck!

**Lil** · October 24th 2010, 09:21 AM

Can I do like this?
$lg\sqrt{x^2 +x-5}=lg1$

**harish21** · October 24th 2010, 09:32 AM

Yes!

Can you complete the problem now?

Hint: What is $\mbox{exp(log(x))}$ ??

**Lil** · October 24th 2010, 09:32 AM

How I solve and try, I can't get D

**Lil** · October 24th 2010, 09:33 AM

Originally Posted by harish21

Yes!

Hint: What is $\mbox{exp(log(x))}$ ??

I don't know... So what it is?

**Jhevon** · October 24th 2010, 09:38 AM

Originally Posted by Lil

Can I do like this?
$log\sqrt{x^2 +x-5}=lg1$

Yes! Now look at the fact i gave you

If log of something = log of something else, then something = something else... hopefully that makes sense when i put it in words

the other two posters are suggesting you plug each side of the equation into the exponential function and equate them. you get to the same place eventually by doing that. it's easier to just drop the logs if you know that fact.

**raa91** · October 24th 2010, 09:38 AM

remember the exponential function is the inverse function of log so think of exp(log x) as ..?

**harish21** · October 24th 2010, 09:38 AM

If log(a) = log(b), then

$exp(log(a))= exp(log(b)) \implies a=b$

apply this to what you have stated in post no 4!

**Lil** · October 24th 2010, 09:43 AM

Originally Posted by Jhevon

Yes! Now look at the fact i gave you

If log of something = log of something else, then something = something else... hopefully that makes sense when i put it in words

the other two posters are suggesting you plug each side of the equation into the exponential function and equate them. you get to the same place eventually by doing that. it's easier to just drop the logs if you know that fact.

Sorry guys, I correct my mistake. Here aren't log, It's just lg.

**Jhevon** · October 24th 2010, 09:45 AM

Originally Posted by Lil

Sorry guys, I correct my mistake. Here aren't log, It's just lg.

The same rules apply

In fact, if you look at the posts of the others, they were assuming it was ln, i wasn't. The fact works in general.

$\displaystyle \log_a X = \log_a Y \implies a^{ \log_a X} = a^{\log_a Y} \implies X = Y$

Here, $\displaystyle a > 1$ is any base. For ln(x), $\displaystyle a = e$ , but as you see, you get the same result. Just drop the logs. You can do this because the logarithm function is 1-1

**e^(i*pi)** · October 24th 2010, 09:46 AM

Originally Posted by Lil

Sorry guys, I correct my mistake. Here aren't log, It's just lg.

It doesn't matter, the rule in posts 2 and 10 holds for all bases except 0 and 1

**Lil** · October 24th 2010, 09:51 AM

Originally Posted by Jhevon

Just drop the logs.

$x^2 +x-5=1$

**Jhevon** · October 24th 2010, 09:53 AM

Originally Posted by Lil

$x^2 +x-5=1$

Yes, now solve that quadratic.

(Always check if your answers are in the domains of the functions with problems like these. And please remember there is a specific reason why you can just "drop" the logs here. It is not true that you can do that with functions in general, only 1 - 1, that is, invertible ones)

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