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Logarithms solution

  1. Oct 20, 2013 #1
    1. The problem statement, all variables and given/known data

    [itex]2^{3x} = 7^{x+1}[/itex]

    2. Relevant equations


    3. The attempt at a solution

    [itex]2^{3x} = 7^{x+1}[/itex]

    I'm not entirely sure how to proceed from here. I'm assuming that the following is the right step towards the solution...

    [itex](log2)(3x) = log(7)(x+1)[/itex]

    ...but I'm not entirely sure.

    Any suggestions?

  2. jcsd
  3. Oct 20, 2013 #2


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    You are doing just fine. Now solve for x.
  4. Oct 20, 2013 #3
    I've arrived at

    [itex]\frac{3x}{x+1} = \frac{log7}{log2}[/itex]

    I'm not sure how to progress.
    What should I do next?

  5. Oct 20, 2013 #4


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    You are kind of going the wrong way. You've got log(2)*3x=log(7)(x+1)=log(7)x+log(7). Move all the terms involving x to one side and solve for x. log(2) and log(7) are just numbers.
    Last edited: Oct 20, 2013
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