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Apply algebraic reasoning to show that a=b^(loga/logb) for any a,b>0

  • Math -

    take the log of each side
    loga=(loga/logb)logb
    now reduce.

  • Math -

    I have:

    loga=(loga/logb)logb
    (loga/logb )(1/loga)=logb
    (1/logb)=logb

    Ok now what?

  • Math -

    you erred.
    loga=(loga/logb)logb
    the logb on the right side divide out (one on numerator, one in denominator)
    loga=loga
    divide both sides by loga.

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