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Solve for X. Leave answers in logarithmic forms.

3^(2-x) = 2^x


  • trig -

    take the logs of both sides
    log (3^(2-x)) = log (2^x)
    (2-x)log3 = xlog2
    2log3 - xlog3 = xlog2
    2log3 = xlog2 + xlog3
    2log3 = x(log2 + log3)
    x = 2log3/(log2 + log3)
    or x = 2log3/log6

  • trig -

    Ok, I'm kind of confused.

    A log's base typically 10 right? Why would you write log 3^(2-x)? Shouldn't it be log(sub3) 2-x) ?

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