Help with logs?

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4 Answers

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  • Favorite Answer

    Do you know how to solve quadratics? Let e^(x) = u

    u^2 - 2u - 24 = 0

    u^2 - 2u = 24

    u^2 - 2u + 1 = 24 + 1

    (u - 1)^2 = 25

    u - 1 = -5 , 5

    u = 1 - 5 , 1 + 5

    u = -4 , 6

    e^(x) = -4 , 6

    There's no real solution for e^(x) = -4. That's extraneous

    e^(x) = 6

    x = ln(6)

  • DWRead
    Lv 7
    2 years ago

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  • 2 years ago

    Think of a logarithm as a reverse exponent. So log(10) 100 = 2 could be rewritten as 10^2 = 100.

  • Steven
    Lv 4
    2 years ago

    Agree with first answer

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