A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
I will fan and medal!!! Help please!!!!
Solve: log (2x + 1) = log (x  1).
anonymous
 one year ago
I will fan and medal!!! Help please!!!! Solve: log (2x + 1) = log (x  1).

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0https://malaysia.answers.yahoo.com/question/index?qid=20111115184243AAGxxTV

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1Well... Whenever you deal with a logarithimic equation your aim is to make all the logarithms have the same base, which in this case is 10, so when that happens you can just deal with what is inside, it looks like this: \[\log_{b} a=\log_{b} c\] Then: \[a=c\] So, for your excercise: \[\log(2x +1)=\log(x1) \] You can see that there's already a logarithm with the same base on both sides, so therefore, we can just focus on what s inside of them: \[2x +1=x1\] And all you have to do is solve for "x"

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Owlcoffee I got x=2/3

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1That's incorrect. You see, when you have: \[2x +1=x1\] And we subtract "x" on both sides, and subtract 1 as well, we end up with: \[x=11\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Muathasim I actually had that as my original answer and it got marked wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you sure the question is exactly the same?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes I just double checked @Muathasim

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It is because the domain of log x is x>0, meaning you cannot take the log of a negative number. When you substitute 2 in for x in either 2x + 1 or x 1, the expression returns a negative value. So the equation has no solution

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Brilliant answer. I did not think of that.

Owlcoffee
 one year ago
Best ResponseYou've already chosen the best response.1Now, here's where it becomes a little complex. We have to take in consideration the range of existance of the logarithms, by looking at the very definition: \[\log_{a} b=c <=>a^c=b\] \[b>0\] \[a>0\] \[a \neq1\] so, if we take the solution calculated and plug it in the excercise: \[\log(2(2)+1)=\log(21)\] \[\log(3)=\log(3)\] Thing is, if you look at the definition, the numbers inside of the logarithm cannot be negative, which in this case they are, so we can conclude this equation has no solution inside the definition of the logarithms.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.