Quote:
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Originally Posted by Scrapper1
What's x so that 2^x yields an answer ending in 0? There isn't any!
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The first of the infinite values of x occurs at 2^x = 10 for values of x > 1.
Use
log2(n) = log(n)/log(2)
= log(n)/0.30103 (approximate to 5 decimal places)
= log(n) x 3.32192
Solving 2^x = 10
log2(10) = log(10) x 3.32192
or x = 3.32192
As I said this is the first of the infinite number of x values satisfying your condition.