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#1) The half life formula is derived from a manipulation of a half life and decay time. In general it is:

Y = N (1/2)^(t/h)

Our information is:
Y = remaining quantity = 50% of 50g = 26.5g C-14
N = initial quantity = 50g C-14
t = time = (what we are solving for...)
h = half-life constant = 5700 years for C-14

26.5 = 50 (1/2)^(t/5700) -- taking logs we have...
ln(26.5) = ln(50(1/2)^(t/5700)) -- using the properties of logs...
ln(26.5) = ln(50) + (t/5700)*ln(1/2) -- solving for t...
t = 5220.83

Our fossil is 5220.83 years old, which makes sense because we have a little more than half of the C-14 remaining, so our age should be a little less than the C-14 half life.

#2) We are solving for e^x + e^-x = 5 algebraically.

e^x + e^-x = 5 -- subtract both sides by five...
e^x + e^-x - 5 = 0 -- multiply our equation by e^x...
e^2x + 1 - 5e^x = 0
e^2x - 5e^x + 1 = 0, let y = e^x
y^2 - 5y + 1 = 0 -- use our quadratic formula...

y = e^x = (5 +/- sqrt 21)/2 -- taking logs...

x = ln((5 +/- sqrt 21)/2) = +/- 1.5568

The second method I originally thought of is just more out of interest, I guess.

I first recognized that for some x, cosh(x) = (e^x + e^-x)/2, so our equation is basically: cosh(x) = 5/2 and we must solve:

x = arccosh(5/2), which has a formula similar to the one before, however this only gives x = 1.5568 as a value due to domain restrictions.

Last edited by pbandjay; September 2nd 2009 at 10:45 PM.