anonymous
  • anonymous
I will fan and medal!!! Help please!!!! Solve: log (2x + 1) = log (x - 1).
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
https://malaysia.answers.yahoo.com/question/index?qid=20111115184243AAGxxTV
Owlcoffee
  • Owlcoffee
Well... 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: \[\log(2x +1)=\log(x-1) \] 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: \[2x +1=x-1\] And all you have to do is solve for "x"
anonymous
  • anonymous
@Owlcoffee I got x=-2/3

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
x = -2
Owlcoffee
  • Owlcoffee
That's incorrect. You see, when you have: \[2x +1=x-1\] And we subtract "x" on both sides, and subtract 1 as well, we end up with: \[x=-1-1\]
anonymous
  • anonymous
@Muathasim I actually had that as my original answer and it got marked wrong
anonymous
  • anonymous
Are you sure the question is exactly the same?
anonymous
  • anonymous
yes I just double checked @Muathasim
anonymous
  • anonymous
It 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
  • anonymous
@peachpi thank you
anonymous
  • anonymous
Brilliant answer. I did not think of that.
Owlcoffee
  • Owlcoffee
Now, 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: \[\log(2(-2)+1)=\log(-2-1)\] \[\log(-3)=\log(-3)\] 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.