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A Logarithm Question?

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Write 2lnx + (1/3)ln5 - 3ln2 as one logarithm.

I feel like I REALLY should be remembering something but I can't pin it down. Help?

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  1. Hi,

    You need the following laws of logarithms:

    log(xy) = log x + log y

    log(x/y) = log x - log y

    log(x^n) = n∙log x

    Therefore,

    2lnx + (1/3)ln5 - 3ln2

    = ln(x^2) + ln(5^1/3) - ln(2^3)

    = ln(5^1/3 ∙ x^2) - ln(8)

    = ln(5^1/3 ∙ x^2 ÷ 8)

    You might want to write 5^1/3 as the cube root of 5 (I just can't type it out).  Hope that helps.

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