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Math Help - Solving equations using Logs

  1. #1
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    Question Solving equations using Logs

    Usually I'm pretty good in math, and I get the concept of logs but i can't figure out how to solve problems like this:

    10^{1-x} = 6^x


    It looks pretty simple, and here is how I start it out:
    1-x = log(6^x

    1-x = x log6

    Then I can't figure out how to isolate the x after that. I'm sure it is just something really simple that I'm missing, but I just can't figure it out.
    Last edited by Kieth89; April 6th 2011 at 10:12 AM.
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  2. #2
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    I think you should fix your formatting a bit. You have

    10^{1-x}=6^{x}. The code for that is 10^{1-x}=6^{x}.

    Taking the log of both sides yields

    \ln(10^{1-x})=\ln(6^{x}), which implies that

    (1-x)\ln(10)=x\ln(6).

    Can you finish?

    [EDIT]: If you use the base-10 logarithm instead of base e, your LHS does simplify a bit.
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  3. #3
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    I still don't see how to finish. You could do this:

    ln(10)/ln(6) = x/(1-x)

    but that just leads to a dead end I think, because there is no way to solve for x without getting rid of the fractions.
    So then I thought about distributing the ln(10) but I'm almost positive that that is breaking a rule. So I'm out of ideas.
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  4. #4
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    Multiplying and dividing aren't the only tools in your toolbox. What about this:

    \ln(10)=x\ln(6)+x\ln(10)?

    What could you do then?
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  5. #5
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    Quote Originally Posted by Kieth89 View Post
    Usually I'm pretty good in math, and I get the concept of logs but i can't figure out how to solve problems like this:

    10^{1-x} = 6^x


    It looks pretty simple, and here is how I start it out:
    1-x = log(6^x

    1-x = x log6

    Then I can't figure out how to isolate the x after that. I'm sure it is just something really simple that I'm missing, but I just can't figure it out.
    I've learnt that it's always a good idea to move over all the x's to one side when doing an equation. If we do it in your example we get this

    1= x log6+x

    When this is done, it's often possible to factor out the x like this:

    x log6+x=x(1+log6)
    Last edited by scounged; April 6th 2011 at 10:52 AM.
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  6. #6
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    Thank you!

    Okay, I've got it figured out now. I don't know why I didn't think about factoring out the x's before. I just get fixated on one idea when doing math sometimes and forget about the other options. Thank you for all the help!
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  7. #7
    A Plied Mathematician
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    You're welcome for my input!
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  8. #8
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    Hi keith89,
    Your last equation is correct but you should have explained what Ackbeet is suggesting you do.Use the log base 10 rather than ln base e
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