number of digit of a number 2^(2016) and 5^(2016) is?


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Question Number 93911 by i jagooll last updated on 16/May/20

$number of digit of a number$ $2^{2016} and 5^{2016} is ?$
Commented by PRITHWISH SEN 2 last updated on 16/May/20

$2^{2016} = \frac{{(2^{10})}^{203}}{2^{14}} = \frac{{(1024)}^{203}}{2^{10} . 2^{4}} ∽ \frac{{(10)}^{609}}{10^{3}} ∽ {(10)}^{606}$ $= 606 + 1 digits = 607 digits$ $I am not sure if it works . So please check .$ $5^{2016} = \frac{10^{2016}}{2^{2016}} = \frac{2017 digits}{607 digits} = 1410 digits$
Commented by i jagooll last updated on 16/May/20

$sir 5^{2016} i got 1410 ? it correct ?$
Commented by Tony Lin last updated on 16/May/20

${l o g}_{10} 2^{2016} = 2016 {l o g}_{10} 2$ $\approx 606.8765$ $\Rightarrow 2^{2016} \approx 10^{606.8765} \approx 7.525 \times 10^{606}$ $\to 607 d i g i t s$ ${l o g}_{10} 5^{2016} = 2016 {l o g}_{10} 5$ $\approx 1409.1235$ $\Rightarrow 5^{2016} \approx 10^{1409.1235} \approx 1.329 \times 10^{1409}$ $\to 1410 d i g i t s$
Commented by PRITHWISH SEN 2 last updated on 16/May/20

$I think 10^{606.8765} = 606 + 1 = 607 digits$ $and 10^{1409.1235} = 1409 + 1 = 1410 digits$ $thank you sir .$
Commented by i jagooll last updated on 16/May/20
what if students don't memorize grades from log base (10) 2 and log base (10) 5 sir?
Commented by i jagooll last updated on 16/May/20

$is there another way ?$
Commented by Tony Lin last updated on 16/May/20

$a c t u a l l y$ $l o g 5 = 1 - l o g 2$ $j u s t r e m e m b e r$ $l o g 2 \approx 0.3010$ $l o g 3 \approx 0.4771$ $l o g 7 \approx 0.8451$ $i n s e n i o r h i g h s c h o o l$ $T e a c h e r s a s k e d u s t o r e m e m b e r t h e s e$ $t h r e e c o m m o n v a l u e s, w h i c h i s e a s y$ $t o e v a l u t e a^{\pm b} i f b i s a l a r g e n u m b e r$ $A n d s o m e t i m e s i t w o u l d g i v e y o u t h e$ $v a l u e i n t h e t e s t, s o {t h e r e}^{'} s n o w o r r y$ $a b o u t i t$ $i n u n i v e r s i t y$ $C a l c u l a t o r s a r e a l l o w e d i n s o m e c o u r s e s$ $A n d l o g a r i t h m i s n o t t h e m a i n t o p i c$ $o f t h e t e s t$ $B u t$ $O n e m e t h o d t o e v a l u a t e l o g a r i t h m$ $I c a m e u p w i t h i s u s i n g T a y l o r s e r i e s$ $l o g 2 = \frac{l n 2}{l n 10}$ $W e k n o w t h a t$ $l n (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \cdot \cdot \cdot$ $f o r - 1 < x ⩽ 1$ $p l u g 1 i n$ $\Rightarrow l n 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \cdot \cdot \cdot \approx 0.693147$ $W e c a n n o t p l u g 9 i n t h e M e r c a t o r s e r i e s$ $b e c a u s e - 1 < x ⩽ 1$ $b u t w e c a n u s e$ $l n (\frac{1 + x}{1 - x})$ $= l n (1 + x) - l n (1 - x)$ $p l u g \frac{9}{11} i n$ $s o l n 10$ $= [\frac{9}{11} - \frac{{(\frac{9}{11})}^{2}}{2} + \frac{{(\frac{9}{11})}^{3}}{3} - \frac{{(\frac{9}{11})}^{4}}{4} + \cdot \cdot \cdot]$ $+ [\frac{9}{11} + \frac{{(\frac{9}{11})}^{2}}{2} + \frac{{(\frac{9}{11})}^{3}}{3} + \frac{{(\frac{9}{11})}^{4}}{4} + \cdot \cdot \cdot]$ $= 2 \times [\frac{9}{11} + \frac{{(\frac{9}{11})}^{3}}{3} + \frac{{(\frac{9}{11})}^{5}}{5} + \cdot \cdot \cdot]$ $= 2 \times \frac{\frac{9}{11}}{1 - {(\frac{9}{11})}^{2}} (1 + \frac{1}{3} + \frac{1}{5} + \cdot \cdot \cdot)$ $\approx 2.302585$ $∴ l o g 2 = \frac{l n 2}{l n 10} \approx 0.3010$
Commented by i jagooll last updated on 16/May/20

$ok sir . thank you$
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