Thread: Power number
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Old 2020-08-02, 06:40   #13
axn
 
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Jun 2003

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Quote:
Originally Posted by retina View Post
Okay. But it still doesn't meet the criterion 99%.

log(6¹⁰⁰)/log(7×6¹⁰⁰+11) = 98.92...%

Indeed for any integer N I can't see how you ever get 99%. The only way would be something like e×e⁹⁹+0 = 99%. So k=e, b=e, n=99, c=0. And other multiples of those values.
Ah! You're thinking 99% as an exact number, whereas I'm pretty sure OP intended to give the general flavor of the structure of the number. Meaning, b^n part is the dominant term.
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