Question

How can we solve (1023/1024)ⁿ = 0.5 for n?

Answer

Definitions:

1. log_ax = N means that a^N=x
.
2. log x means log₁₀ x. All log_a rules apply for log. When a logarithm is written without a base it means common logarithm.
3. 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.

1. Inverse properties: log_a a^x = x and a^{(log_a x)} = x
2. Product: log_a (xy) = log_a x + log_a y
3. Quotient: log_a (x/y) = log_a x - log_a y
4. Power: log_a (x_p) = p log_a x
5. Change of base formula: log_a x = log_bx/log_ba

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/-4.2432292765179442545697262837499e-4
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 7.3361399919156076009302055016313e-50
= 3.6680699959578038004651027508157e-50

So the probability of failure is 1-3.6680699959578038004651027508157e-50, 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/-4.3429448190325182765112891891682e-31
= 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 1-3.6680699959578038004651027508157e-50 causes n to be understated by a factor of about 21.