Can not find site to log on for e-learning - Computers & Internet

First, I'm not a facebook representative.
* This is from their site;
I can't log in.

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https://www.facebook.com/help/105487009541643?helpref=topq

Aloha, ukeboy57

open their site and start the sign up procedure

Assuming you are somehow connected to the internet, simply point your web browser at Welcome to Facebook Log In Sign Up or Learn More and log in.

Go to this site and sign up following the instructions there :
https://www.facebook.com/

The web browser you are using may not support flash that the site may be

Point your web browser at Welcome to Facebook Log In Sign Up or Learn More

Facebook is not a program that is installed on your PC... it is a web site... just go to Welcome to Facebook Log In Sign Up or Learn More and log in....

This is not a calculator problem. Use the rules you have learned to simplify your problem.
For a a log in any base. log(a*b) =log(a) +log(b) or log(a)+log(b) =log(a*b).
Thus
log(x+1)+log(x-2)= log[(x+1)*(x-2)]=1
Use the fact that log4=log in base 4 and rewrite the equation with the appropriate symbols. log4[(x+1)*(x-2)] =1
However 1=log4(4) and
log4[(x+1)*(x-2)]=log4(4)
The equality of the two logs implies the equality of their arguments (contents) and

(x+1)*(x-2) =4

Now you solve this quadratic equation by the methods you must have learned (sorry I have to leave some thing for you to do, to ensure that proper learning is achieved).
Once you find the roots of the quadratic equation, verify that each term in your original expression has meaning-- x+1 positive and x-2 positive.

If one of the roots makes the argument (x+1) or (x-2) negative, reject it. because the argument of a log function cannot be negative.