Question
How can we solve (1023/1024)^{n} = 0.5 for n?
Answer
What we have here is an exponential equation, where the variable, n, shows up in the exponent. Solving exponential equations
involves logarithms.
Definitions:
 log_{a}x = N means that a^{N}=x
.
 log x means log_{10} x. All log_{a} rules apply for log.
When a logarithm is written without a base it means common logarithm.
 ln x means log_{e} x, where e is about 2.718. All log_{a}
rules apply for ln. When a logarithm is written "ln" it means natural logarithm.
Note: ln x is sometimes written Ln x or LN x.
Rules:
 Inverse properties: log_{a} a^{x} = x and a^{(loga x)} = x
 Product: log_{a} (xy) = log_{a} x + log_{a} y
 Quotient: log_{a} (x/y) = log_{a} x  log_{a} y
 Power: log_{a} (x_{p}) = p log_{a} x
 Change of base formula: log_{a} x = log_{b}x/log_{b}a
Using these definitions and rules, we take the logarithm of both sides of the given equation. So:
n log(1023/1024) = log 0.5
n = log 0.5/log(1023/1024)
Using the log function on a calculator:
n = 0.30102999566398119521373889472449/4.2432292765179442545697262837499e4
n = 709.43608286708182323352062206674
Applying this to the jigsaw puzzle example from Getting Improbable Results
the probability of success in a single toss including the first piece landing picture side up is:
1/2 x (1/2 x 1/1440 x 1/100)^{9}
= 1/2 x (1/288000)^{9}
= 1/2 x 7.3361399919156076009302055016313e50
= 3.6680699959578038004651027508157e50
So the probability of failure is 13.6680699959578038004651027508157e50, but this number is too close to 1
for the computer to tell the difference so we substitute the number less than and closest to 1 that the
computer can handle. That is 0.999999999999999999999999999999. This hughely decreases the chance of failure
but does give us a conservative estimate of the number of tosses to reach a 50% chance of success. So:
n log 0.999999999999999999999999999999 = log 0.5 is used to calculate a conservative n.
n = 0.30102999566398119521373889472449/4.3429448190325182765112891891682e31
= 693147180559945309417232121457.83
Of course in the Observation essay I rounded off these very long numbers produced by the calculator.
In fact the conservatism induced by using 0.999999999999999999999999999999 as a substitute for 13.6680699959578038004651027508157e50
causes n to be understated by a factor of about 21.
