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Math Help - Logarithm inequality with different constants?

  1. #1
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    Logarithm inequality with different constants?

    The problem is as such:

    5*3x>12*5x

    I have been mulling over this thing for quite some time now. Normally I would start by dividing both sides by the 5, but that doesnt isolate the x, so that doesnt seem to be the way to go. So I turned everything into logs.

    log 5 + x log 3 > log 12 + x log 5

    From here, I have tried just about everything I could come up with. Subtract around so that you get the x logs on one side, and the non-variable logs on the other side, but then they dont seem to match up. Can someone let me know where I go from here? I am sketchy on what is allowed with this. Can I simply have x log 3 - x log 5, and combine them somehow, even though one is a log 3, and the other is a log 5? I feel like my head is swimming from all the conversions back and forth between log and not log formats.
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  2. #2
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    Re: Logarithm inequality with different constants?

    Quote Originally Posted by Latsabb View Post
    log 5 + x log 3 > log 12 + x log 5
    In this equation log is to the same (arbitrary) base. This is a linear equation in x.
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  3. #3
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    Re: Logarithm inequality with different constants?

    from  log10(5)+xlog10(3)>log10(12)+xlog10(5)
     xlog10(3)-xlog10(5)>log10(12)-log10(5)
    x(log10(3)-log10(5))>log10(12)-log10(5)
    x<(log10(12)-log10(5))/(-log10(5)-log(3))
    x<-(log10(12/5)/(log10(5/3))
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  4. #4
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    Re: Logarithm inequality with different constants?

    Ok, so then I break that down to

    x log 3/5 > log 12/5

    So I could then isolate x by dividing by log 3/5 on both sides, and flipping the inequality sign. However, the book states that the answer must be in "exact" form. (ie log 2/3, not 0.63, for example)

    Just to make sure, I can just call that (log 12 - log 5) / (log 3 - log 5) correct? That seems like it should be ok in my head, but like I said, I have been mulling over things for a while, and I am kind of burnt.
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  5. #5
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    Re: Logarithm inequality with different constants?

    Personally, I would prefer x> \frac{log(12/5)}{log(3/5} rather than using the differences. Note the ">" rather than "<". Since 3/5< 1, log(3/5) is negative and dividing both sides by a negative number reverses the direction of the inequality.
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  6. #6
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    Re: Logarithm inequality with different constants?

    But since the problem originated with the >, shouldnt it be flipped to <?

    Also, how would this be worked out if there was a negative exponent of x? Lets say:

    3x - 4 * 3-x > 0

    It seems as if I put those into logs, it would turn into

    x log 3 + (-x)log 3 > log 4

    Would that be the correct way of doing it? Because that looks like the two x log 3 groups should cancel each other out. Even if I use x log 1/3, wouldnt that then be x log (3)(1/3) and just become log 1, which would be log 0?
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  7. #7
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    Re: Logarithm inequality with different constants?

    Hello, Latsabb!

    \text{Solve for }x\!:\;\;5\cdot3^x\:>\:12\cdot5^x

    Take logs, the base doesn't matter . . . Let's use natural logs.

    We have: . \ln(5\cdot3^x) \;>\;\ln(12\cdot5^x)

    . . . . . \ln(5) + \ln(3^x) \;>\;\ln(12) + \ln(5^x)

    n . . . \ln(5) + x\ln(3) \;>\;\ln(12) + x\ln(5)

    . . . x\ln(3) - x\ln(5) \;>\;\ln(12) - \ln(5)

    . . . x\big[\ln(3) - \ln(5)\big] \;>\;\ln(12) - \ln(5)

    n . . . . . . . . x\ln(\tfrac{3}{5}) \;>\;\ln(\tfrac{12}{5})


    Divide by \ln(\tfrac{3}{5}) . . . Note that it is negative.

    . . . . . . . . . . . . . x \;{\color{red}<} \;\frac{\ln(\frac{12}{5})}{\ln(\frac{3}{5})}

    . . . . Therefore: . x \;<\;\text{-}1.713830898
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  8. #8
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    Re: Logarithm inequality with different constants?

    Hello again, Latsabb!

    You should NOT take logs of a sum or difference.
    It usually leads nowhere.



    How would this be worked out if there was a negative exponent of x?

    Let's say: . 3^x - 4\cdot3^{-x} \;>\; 0

    Multiply by 3^x\!:\;\;3^{2x} - 4 \;>\;0 \quad\Rightarrow\quad 3^{2x} \;>\;4

    Take logs: . \ln(3^{2x}) \;>\;\ln(4) \quad\Rightarrow\quad 2x\ln(3) \;>\;\ln(4)

    Therefore: . x \;>\;\frac{\ln(4)}{2\ln(3)} \;\approx\;0.630929754
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