<PRE><B>Question 821092</B>
How do i solve A=ln2 in terms of A
2 = e^A
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ln(16) = ln[(2^4)] = ln[(e^A)^4] = ln[e^(4A)] = 4A
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ln(square root of 2) = ln[2^(1/2)] = (1/2)ln(2) = (1/2)ln[e^A] = (1/2)A
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ln(1/2) = ln[2^-1] = ln[(e^A)^-1] = ln[e^(-A)] = -A
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Cheers,
Stan H.
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