if h(x) = ln(2-3x)^5, find H (0)?

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  • 1 month ago

    It is impossible to find H(0) from and expression given as h(x). The capital letter and small letter changes the whole thing. Unless if it was just a typo, then

    H(0) or h(0) (whichever one is correct) 

    = ln[2-3(0)]^5  (just replace x with 0)

    = ln2^5

    = 5ln2 (law of logarithm puts 5 here) 

    ≈ 5 × 0.693

    = 3.465

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  • 1 month ago

    h(x) = ln(2 - 3x)^5

    h(0)

    = ln(2)^5

    = 0.160002698            

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  • MyRank
    Lv 6
    1 month ago

    h(x) = ln(2-3x)^5

    h(0) = ?

     h(x) = 5ln(2-3x)

    put x = 0

    h(0) = 5ln(2-3(0))

    h(0) = 5ln(2)

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  • 1 month ago

    If that's the logarithm of the whole expression:

       h(x) = ln [(2 - 3x)^5] = 5 ln (2 - 3x)

    ...then h(0) = 5 ln 2 ~~ 3.4657

    If that's the logarithm raised to the 5th power:

        h(x) = [ln (2 - 3x)] ^ 5      .... also written as log^5 (2 - 3x)

    ...then h(0) = (ln 2)^5 ~~ 0.16000

    The second interpretation is the one most commonly used in typewritten expressions when using the function notation as in "ln(x): instead of using ln as an operator as in "ln x".

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  • 1 month ago

    Presuming that is:

    h(x) = ln[(2 - 3x)⁵]

    Which would simplify to:

    h(x) = 5 ln(2 - 3x)

    Solve for h(0):  substitute 0 for x and simplify:

    h(0) = 5 ln(2 - 3 * 0)

    h(0) = 5 ln(2 - 0)

    h(0) = 5 ln(2)

    or approximately:

    h(0) = 5(0.693147)

    h(0) = 3.465735

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  • 1 month ago

    I have no idea sorry

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