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    Hey guys I need help solving these 2 questions.

    1) Solve 3^x = 4^2-x.

    So here I have got to the stage where I am at:
    2x log 3 = (2-x) log 4

    What do I do now?

    And the second question is to solve this for x:

    2 log (2x) = 1 + log a

    I don't know where to start on this one

    Any thoughts? Thanks
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    For the first one you should have x log 3 = (2-x) log 4.
    Expand the RHS so you get 2log4 - xlog4
    Bring the xs to one side so you get xlog3 + xlog4 = 2log4
    Bring x outside and divide through should give you the answer.

    For the second equation I bought the 2 inside the log to give me log(4x^2) = 1 + log a
    I rewrote the RHS as log 10 + log a, and then as log 10a.
    So we have log(4x^2) = log (10a)
    All you need to do then is get rid of the logs and solve.
 
 
 
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