
#1
Feb2311, 09:16 AM

P: 273

I was playing around with my calculator and I noticed that to calculate the number of digits in a number n we have # of digits in n = [log(n) + 1], where log is in base 10 and [] represents the floor function. I after looked this up and it seems like its true. But how would you go about proving something like this? I thought of writing n as its decimal expansion, and then taking the logarithm, but it doesn't seem to get me no where. Any ideas?




#2
Feb2311, 10:00 AM

Mentor
P: 16,690

Let x be a number, then it is easy to see that
x has n digits if and only if [tex]10^{n1}\leq x<10^{n}[/tex]. If we apply your expression on this, then we obtain [tex]n=\log(10^{n1})+1\leq \log(x)+1<\log(10^n)+1=n+1[/tex] This yields that [tex][\log(x)+1]=n[/tex]. 



#3
Feb2311, 10:01 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,895

A number, n, has "d" digits if and only if [itex]10^{d1}\le n< 10^d[/itex]. Take the logarithm (base 10) of each part of that, using the fact that logarithm is an increasing function.




#4
Feb2311, 10:13 AM

P: 273

Number of digits in a number?
Ahhh didn't think of that. Thanks a lot to both of you! :)




#5
Feb2311, 07:41 PM

P: 828





#6
Feb2311, 09:27 PM

P: 273




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