Find the number of digit in 2^(50)


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Question Number 62449 by Tawa1 last updated on 21/Jun/19

$Find the number of digit in 2^{50}$
Commented by Tawa1 last updated on 21/Jun/19

$And can we find a general nth number of term in any number and powers$
	Answered by Rasheed.Sindhi last updated on 21/Jun/19

	$B y e x p e r i m e n t i n g b e l o w w e c a n s e e$ $t h a t f i r s t t h r e e p o w e r s o f 2 a r e 1 - d i g i t$ $n e x t t h r e e a r e 2 - d i g i t, a f t e r t h a t$ $n e x t t h r e e a r e 3 - d i g i t n u m b e r s . . .$ $[\begin{matrix} 2^{m} & N & d \\ 2^{1} & 2 & 1 \\ 2^{2} & 4 & 1 \\ 2^{3} & 8 & 1 \\ 2^{4} & 16 & 2 \\ 2^{5} & 32 & 2 \\ 2^{6} & 64 & 2 \\ 2^{7} & 128 & 3 \\ 2^{8} & 256 & 3 \\ 2^{9} & 512 & 3 \\ 2^{10} & 1024 & 4 \\ . . . & . . . & . . . \end{matrix}]$ $W e o b s e r v e t h a t :$ $2^{3 n} h a v e n d i g i t$ $2^{3 n - 1}, 2^{3 n - 2} a l s o h a v e n d i g i t$ $N o w,$ $2^{51} = 2^{3 \times 17} h a v e 17 d i g i t s$ $2^{50} = 2^{3 \times 17 - 1} a l s o h a v e 17 d i g i t s$
	Commented by mr W last updated on 21/Jun/19

	$d e a r r a s h e e d s i r, t h e i d e a i s g r e a t!$ $b u t t h e n u m b e r o f (d e c i m a l) d i g i t s$ $o f 2^{n} {d o e s n}^{'} t i n c r e a s e i n a c o n s t a n t$ $s t e p o f 3 a s y o u d e s c r i b e d . y o u h a v e$ $h a d s e e n t h i s i f y o u h a d a c o u p l e$ $e x a m p l e s m o r e, e . g .$ $2^{10} = 1024$ $2^{11} = 2048$ $2^{12} = 4096$ $2^{13} = 8192 \Rightarrow 4 n u m b e r s w i t h 4 d i g i t s!$ $2^{14} = 16384$ $2^{15} = 32768$ $2^{16} = 65536 \Rightarrow 3 n u m b e r s w i t h 5 d i g i t s$ $2^{17} = 131072$ $. . . . . .$ $i n f a c t 2^{50} = 11258 99906 84262 4$ $i t h a s 16 d i g i t s, n o t 17 .$
	Commented by Rasheed.Sindhi last updated on 21/Jun/19

	$S i r t h a n k s f o r g u i d a n c e . I l e a r n t$ $t h a t o n e s h o u l d b e m o r e c a r e f u l$ $i n h a n d l i n g n u m b e r s!$ $U p t o 2^{30} t h e r u l e^{'} n u m b e r o f d e c i m a l$ $d i g i t s o f 2^{3 n} = n^{'} w o r k s w e l l, b u t a t$ $2^{33} t h e r u l e {d o e s n}^{'} t w o r k!$ $T h a n k S a g a i n s i r!$
	Commented by mr W last updated on 21/Jun/19

	$d e a r s i r, i a p p r e c i a t e y o u r d e e p t h o u g h t s$ $v e r y m u c h .$
	Answered by mr W last updated on 22/Jun/19

	Commented by Rasheed.Sindhi last updated on 22/Jun/19

	$N i c e A p p r o a c h S i r!$ ${T h a t}^{'} s a c c o r d i n g t o l o g i c a l t h i n k i n g!$ $A n d {t h a t}^{'} s d e e p t h i n k i n g a c t u a l l y!$ $(W h e r e a s I t h o u g h t i n m y a n s w e r p a t t e r n i c a l l y, i n a b l i n d w a y .)$ $B T W S i r,$ $⌊ n \log 2 ⌋ + 1 \overset{?}{=} ⌈ n \log 2 ⌉$ $(R e c a l l t h a t n \log 2 {i s n}^{'} t a w h o l e n u m b e r)$
	Commented by mr W last updated on 22/Jun/19

	$s i m i l a r l y t h e n u m b e r o f d i g i t s o f 3^{n}$ $i s ⌊ n \log 3 ⌋ + 1 .$
	Commented by mr W last updated on 22/Jun/19

	$t h a n k s s i r!$ $y o u a r e a b s o l u t e l y c o r r e c t w i t h$ $⌊ n \log 2 ⌋ + 1 = ⌈ n \log 2 ⌉ s i n c e \log 2 i s$ $n o t a r a t i o n a l n u m b e r . a c t u a l l y i$ $a l s o c o n s i d e r e d t o u s e ⌈ n \log 2 ⌉ i n s t e a d$ $o f ⌊ n \log 2 ⌋ + 1, b u t m y p e r s o n a l t a s t e$ $i s m o r e f o r f l o o r f u n c t i o n .$
	Commented by Rasheed.Sindhi last updated on 22/Jun/19
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