Question 413440
I'm not sure, but I think this is what you wrote:
{{{b = ln(e^t) - ln(e^D)}}}
If so,
{{{b = ln(e^t/e^D)}}}
{{{b = ln(e^(t - D))}}}
{{{b = t - D}}}
{{{t = b + D}}}
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{{{I = log(2,1/p)}}}
{{{I = log(2,p^-1)}}}
{{{I = -log(2,p)}}}
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This is the same concept as this easier problem:
{{{K = log(1/100)}}} (log to the base 10)
{{{K = log(10^-2)}}}
{{{K = -2*log(10)}}}
{{{K = -2}}}