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

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• 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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• h(x) = ln(2 - 3x)^5

h(0)

= ln(2)^5

= 0.160002698

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• 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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• 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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• 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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• I have no idea sorry

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