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anonymous

  • one year ago

I will fan and medal!!! Help please!!!! Solve: log (2x + 1) = log (x - 1).

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  1. anonymous
    • one year ago
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    https://malaysia.answers.yahoo.com/question/index?qid=20111115184243AAGxxTV

  2. Owlcoffee
    • one year ago
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    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"

  3. anonymous
    • one year ago
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    @Owlcoffee I got x=-2/3

  4. anonymous
    • one year ago
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    x = -2

  5. Owlcoffee
    • one year ago
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    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: \[x=-1-1\]

  6. anonymous
    • one year ago
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    @Muathasim I actually had that as my original answer and it got marked wrong

  7. anonymous
    • one year ago
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    Are you sure the question is exactly the same?

  8. anonymous
    • one year ago
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    yes I just double checked @Muathasim

  9. anonymous
    • one year ago
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    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

  10. anonymous
    • one year ago
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    @peachpi thank you

  11. anonymous
    • one year ago
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    Brilliant answer. I did not think of that.

  12. Owlcoffee
    • one year ago
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    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.

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