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
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
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
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.
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 \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
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)