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Math Help - logarithm equation

  1. #1
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    logarithm equation

    Hello, I have these two equations but are really stumped as to how to solve them.

    ln2 + ln(x+3) = 3

    log3 + log(2x-1) = log2 + log(x+1)

    I know the concept of multiplying when you have addition, and division when subtracting. In these two cases, I think I would be multiplying the terms as they are added ?

    Can someone please get me started on these so I can see how to go about it.

    Thanks kindly for any help.
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  2. #2
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    Hint: Simplify everything first using \displaystyle \log{(a)} + \log{(b)} = \log{(ab)}.
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    not sure if I am on the right track, but I have factored the two to:

    ORIGINAL: log3 + log(2x-1) = log2 + log(x+1)
    FACTORED: log3(2x-1) = log2(x+1)

    ORIGINAL: ln2 + ln(x+3) = 3
    FACTORED: ln2(x+3) = 3

    am I correct ?
    If so, what is the next step ?

    Thanks kindly for any help.
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  4. #4
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    Quote Originally Posted by fran1942 View Post
    FACTORED: log3(2x-1) = log2(x+1)
    You can cancel the logs off both sides leaving

    3(2x-1) = 2(x+1)

    Solve for x.
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  5. #5
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    thanks, I get 5/4 for the first one.

    However how can I finish off the second problem:
    Do I have to balance the sides first? I dont think I can just drop off the ln on the left.

    ORIGINAL: ln2 + ln(x+3) = 3
    FACTORED: ln2(x+3) = 3
    ANSWER: ?
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  6. #6
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    Quote Originally Posted by fran1942 View Post
    not sure if I am on the right track, but I have factored the two to:

    ORIGINAL: log3 + log(2x-1) = log2 + log(x+1)
    FACTORED: log3(2x-1) = log2(x+1)

    ORIGINAL: ln2 + ln(x+3) = 3
    FACTORED: ln2(x+3) = 3

    am I correct ?
    If so, what is the next step ?

    Thanks kindly for any help.
    They're both correct. Now you need to exponentiate both sides of both equations.
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  7. #7
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    Quote Originally Posted by Prove It View Post
    They're both correct. Now you need to exponentiate both sides of both equations.
    thanks, would I be correct by doing this:

    ORIGINAL: ln2 + ln(x+3) = 3
    FACTORED: ln2(x+3) = 3
    ANSWER:
    i) e^3 = 2(x+3)
    ii) e^3 = 2x+6
    iii) ((e^3) /2)-6 = x
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  8. #8
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    Quote Originally Posted by fran1942 View Post
    FACTORED: ln2(x+3) = 3
    ANSWER:
    i) e^3 = 2(x+3)
    ii) e^3 = 2x+6
    iii) ((e^3) /2)-6 = x

    Nope, you subtract the 6 first then divide by 2.

    Do you know why? Its really important you understand this concept!
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  9. #9
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    Not quite...

    2x + 6 = e^3
    2x = e^3 - 6
    x = (e^3 - 6)/2
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