mathmath333 one year ago Question

1. mathmath333

\large \color{black}{\begin{align} \normalsize \text{Find the number of digits in }\ 60^{12}\hspace{.33em}\\~\\ \end{align}}

2. mathstudent55

|dw:1438031157075:dw|

3. mathmath333

but calclulting whole stuff is cumbersome

4. ganeshie8

dumb q, but are you allowed to use calculator for taking logarithm

5. mathmath333

no calculo tor

6. ganeshie8

i mean my question of asking whether you're allowed to use a calculator is dumb :)

7. mathmath333

i mean no cheating stuff allowed

8. ganeshie8

number of digits equals below expression, i don't see an easy way to simplify it though$\large \lfloor \log_{10} 60^{12}\rfloor+1$

9. mathmath333

where did u get this formula

10. mathmath333

what is the name of the formula

11. ganeshie8

Its not a formula as such, just work few examples till you get convinced, keeping in mind the definition of logarithm $\large 10^x = N ~~\iff~~ x=\log_{10} N$

12. mathmath333

13. ganeshie8

how many digits are there in $$\large 10^1$$ ?

14. mathmath333

$$\large \lfloor \log_{10} 60^{12}\rfloor\color{red}{+1}$$

15. mathmath333

2

16. ganeshie8

how many digits are there in $$\large 10^2$$ ?

17. mathmath333

3

18. mathmath333

i see

19. ganeshie8

how many digits are there in $$\large 10^{300}$$ ?

20. mathmath333

301

21. ganeshie8

how did you get 301 ?

22. mathmath333

23. ganeshie8

You got 301 by adding 1 to the logarithm, 300.

24. ganeshie8

It helps to keep in mind the actual definition of logarithm : logarithm = exponent

25. ganeshie8

$\large 10^{\color{red}{300}}$ $$\color{red}{300}$$ is called the logarithm of $$10^{\color{red}{300}}$$ relative to base $$10$$

26. ganeshie8

as you can see, the number of digits in the decimal expansion of $$10^{\color{red}{300}}$$ equals $\large \lfloor \log_{10} 10^{\color{red}{300}}\rfloor + 1$

27. mathmath333

that was easy

28. mathmath333

$$log _{10} 10$$

29. ganeshie8

$$\large 10 = 10^\color{red}{1}$$ so the logarithm of $$10$$ is $$\color{red}{1}$$ (relative to base 10)

30. mathmath333

is it important to include floor function

31. ganeshie8

how many digits are there in $$\large 3^4$$ ?

32. mathmath333

2

33. ganeshie8

how did u get 2 ?

34. ganeshie8
35. mathmath333

81=3^4

36. ganeshie8

try using the formula

37. ganeshie8

$\large \lfloor \log_{10} 3^{4}\rfloor+1$

38. ganeshie8

u will see why floor function is necessary

39. mathmath333

ok

40. ganeshie8
41. mathmath333

it seems i need to remember log values to use the formula

42. ganeshie8

not worth it, there must be a simpler way to solve it

43. ganeshie8

Remember, if you're struggling too hard, then it means only one thing : your method is wrong/not clever

44. mathmath333

ok thnx

45. ganeshie8

is this really supposed to be solved w/o calculator ?

46. mathmath333

yes lol

47. ganeshie8

i don't really see how one computes it without knowing the value of $$\log_{10}6$$, or without working the actual number hmm

48. mathmath333

log6=log2+log3

49. mathmath333

i just need to rote log 2 and log 3

50. ganeshie8

i got better things to memorize than these log values haha

51. mathmath333

its matter of pattern , not rote learning

52. mathmath333

log 2=0.30 , log3=0.47 easy

53. ganeshie8

Okay, then lets pretend that we have access to log2 and log3 values :)

54. ganeshie8

log6 = log2 + log3 = 0.30 + 0.47 = 0.77 ?

55. mathmath333

yep correct

56. ganeshie8

good, plug them in the formula and see..

57. mathmath333
58. ganeshie8

\large{\begin{align} \lfloor \log_{10} 60^{12}\rfloor+1 &= \lfloor 12*\log_{10} 60\rfloor+1\\~\\ &= \lfloor 12*(\log_{10}10 +\log_{10} 6)\rfloor+1\\~\\ &\approx \lfloor 12*(1 +0.77)\rfloor+1\\~\\ &=21+1 \end{align}}

59. mathmath333

http://openstudy.com/users/mathmath333#/updates/55b3954fe4b071e6530e16e9 one more method for finding log

60. ganeshie8

Ahh you wanto use taylor series is it $\ln(1+x)=-\sum_{n=1}^\infty\frac{(-x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$

61. mathmath333

i just will remember values to save time

62. ganeshie8

do let me know if/when you find a better method

63. mathmath333

yep .

64. phi

you can make some reasonable approximations to do this problem $60^{12}= 2^{12} \cdot 3^{12} \cdot 10^{12}$ 2^10 is 1024 or close to 10^3 3^12 is 81^3 and 81 is 2^3 * 10 thus $$3^{12} \approx \left(2^3 \cdot 10^1\right)^3 = 2^9 \cdot 10^3$$ putting this mess together $2^{12} \cdot 3^{12} \cdot 10^{12} \\ 10^3 \cdot 2^2 \cdot 2^9 \cdot 10^3 \cdot 10^{12} \\ 2 \cdot 10^3 \cdot 10^{18} \\ 2 \cdot 10^{21}$ which means 22 digits.