math: pre-calculus

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Write the expression as a single logarithm. Ln(e^3y)+ln(ey)-4

  • math: pre-calculus -

    by definition,
    e^(ln x) = x
    ln(e^x) = x

    ln and e^ are inverse functions, just like x^2 and √x

    Assuming no typos, I see

    ln(e^3y)+ln(ey)-ln(e^4) = ln(e^3y * ey / e^4) = ln(y^2) = 2ln(y)

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