How can we solve (1023/1024)n = 0.5 for n?
What we have here is an exponential equation, where the variable, n, shows up in the exponent. Solving exponential equations
- logax = N means that aN=x.
- log x means log10 x. All loga rules apply for log.
When a logarithm is written without a base it means common logarithm.
- ln x means loge x, where e is about 2.718. All loga
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.
- Inverse properties: loga ax = x and a(loga x) = x
- Product: loga (xy) = loga x + loga y
- Quotient: loga (x/y) = loga x - loga y
- Power: loga (xp) = p loga x
- Change of base formula: loga x = logbx/logba
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
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
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.