1. ## 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

2. 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

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

From here I stuck!

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

5. Yes!

Can you complete the problem now?

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

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

7. Originally Posted by harish21
Yes!

Hint: What is $\mbox{exp(log(x))}$??
I don't know... So what it is?

8. 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.

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

10. 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!

11. 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.

12. 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

13. 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

14. Originally Posted by Jhevon
Just drop the logs.
$x^2 +x-5=1$

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