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Was just doing some Calculus homework in which I got the answer to be:
[0.5e^(2ln2)] - 0.5
What property do I use to simplify e^(2ln2)?
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I checked in the formula's section of this forum
and this wasn't in it explicitly, but might be
implied from 2 other equations.
This MIGHT be right:
Here are two formulas similar to those I found.
n ln(m) = ln(m^n)
where m may be restricted to values above zero??
e^(ln(s)) = s, and s may have to be above zero??
Last edited by John E. Franklin (2008-01-14 17:43:51)
igloo myrtilles fourmis
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I see it in a slightly different way, but I get the same answer as John.
Using laws of powers, e^(2ln2) = e^(ln2 x 2) = (e^ln2)^2.
Clearly, e^ln2 = 2, and so the answer is 2^2 = 4.
Why did the vector cross the road?
It wanted to be normal.
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which continuing gives 1.5 for the full expression
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