I have this one. It's looks easy to solve. I'm trying to solve and nothing. I can't get an answer.

$\displaystyle \frac{1}{2}lg(x^2 +x-5)=lg5x+lg\frac{1}{5x}$

...

$\displaystyle \frac{1}{2}lg(x^2 +x-5)=lg1$ Can I divide /:1/2

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- Oct 24th 2010, 09:09 AMLilLogarithm Equation
I have this one. It's looks easy to solve. I'm trying to solve and nothing. I can't get an answer.

$\displaystyle \frac{1}{2}lg(x^2 +x-5)=lg5x+lg\frac{1}{5x}$

...

$\displaystyle \frac{1}{2}lg(x^2 +x-5)=lg1$ Can I divide /:1/2 - Oct 24th 2010, 09:14 AMJhevon
You need two rules and one fact here:

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

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

Secondly, the rules:

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

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

What can you do with those? I'd apply the rules first before the fact - Oct 24th 2010, 09:17 AMLil
I know this rules... I think, I something do wrong.

$\displaystyle lg(x^2 +x-5)^{\frac{1}{2}}=lg1$

From here I stuck! - Oct 24th 2010, 09:21 AMLil
Can I do like this?

$\displaystyle lg\sqrt{x^2 +x-5}=lg1$ - Oct 24th 2010, 09:32 AMharish21
Yes!

Can you complete the problem now?

Hint: What is $\displaystyle \mbox{exp(log(x))}$?? - Oct 24th 2010, 09:32 AMLil
How I solve and try, I can't get

*D* - Oct 24th 2010, 09:33 AMLil
- Oct 24th 2010, 09:38 AMJhevon
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, 09:38 AMraa91
remember the exponential function is the inverse function of log so think of exp(log x) as ..?

- Oct 24th 2010, 09:38 AMharish21
If log(a) = log(b), then

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

apply this to what you have stated in post no 4! - Oct 24th 2010, 09:43 AMLil
- Oct 24th 2010, 09:45 AMJhevon
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 \displaystyle \log_a X = \log_a Y \implies a^{ \log_a X} = a^{\log_a Y} \implies X = Y $

Here, $\displaystyle \displaystyle a > 1$ is any base. For ln(x), $\displaystyle \displaystyle a = e$, 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, 09:46 AMe^(i*pi)
- Oct 24th 2010, 09:51 AMLil
- Oct 24th 2010, 09:53 AMJhevon
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)