# Logarithm Equation

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• Oct 24th 2010, 10:09 AM
Lil
Logarithm Equation
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
• Oct 24th 2010, 10:14 AM
Jhevon
Quote:

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
• Oct 24th 2010, 10:17 AM
Lil
I know this rules... I think, I something do wrong.
$lg(x^2 +x-5)^{\frac{1}{2}}=lg1$

From here I stuck!
• Oct 24th 2010, 10:21 AM
Lil
Can I do like this?
$lg\sqrt{x^2 +x-5}=lg1$
• Oct 24th 2010, 10:32 AM
harish21
Yes!

Can you complete the problem now?

Hint: What is $\mbox{exp(log(x))}$??
• Oct 24th 2010, 10:32 AM
Lil
How I solve and try, I can't get D
• Oct 24th 2010, 10:33 AM
Lil
Quote:

Originally Posted by harish21
Yes!

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

I don't know... So what it is?
• Oct 24th 2010, 10:38 AM
Jhevon
Quote:

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 :p

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.
• Oct 24th 2010, 10:38 AM
raa91
remember the exponential function is the inverse function of log so think of exp(log x) as ..?
• Oct 24th 2010, 10:38 AM
harish21
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!
• Oct 24th 2010, 10:43 AM
Lil
Quote:

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 :p

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.
• Oct 24th 2010, 10:45 AM
Jhevon
Quote:

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

The same rules apply :p

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
• Oct 24th 2010, 10:46 AM
e^(i*pi)
Quote:

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
• Oct 24th 2010, 10:51 AM
Lil
Quote:

Originally Posted by Jhevon
Just drop the logs.

$x^2 +x-5=1$
• Oct 24th 2010, 10:53 AM
Jhevon
Quote:

Originally Posted by Lil
$x^2 +x-5=1$