mathmath333
  • mathmath333
Question
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} \normalsize \text{Find the number of digits in }\ 60^{12}\hspace{.33em}\\~\\ \end{align}}\)
mathstudent55
  • mathstudent55
|dw:1438031157075:dw|
mathmath333
  • mathmath333
but calclulting whole stuff is cumbersome

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More answers

ganeshie8
  • ganeshie8
dumb q, but are you allowed to use calculator for taking logarithm
mathmath333
  • mathmath333
no calculo tor
ganeshie8
  • ganeshie8
i mean my question of asking whether you're allowed to use a calculator is dumb :)
mathmath333
  • mathmath333
i mean no cheating stuff allowed
ganeshie8
  • ganeshie8
number of digits equals below expression, i don't see an easy way to simplify it though\[\large \lfloor \log_{10} 60^{12}\rfloor+1\]
mathmath333
  • mathmath333
where did u get this formula
mathmath333
  • mathmath333
what is the name of the formula
ganeshie8
  • ganeshie8
Its not a formula as such, just work few examples till you get convinced, keeping in mind the definition of logarithm \[\large 10^x = N ~~\iff~~ x=\log_{10} N\]
mathmath333
  • mathmath333
how did u add +1
ganeshie8
  • ganeshie8
how many digits are there in \(\large 10^1\) ?
mathmath333
  • mathmath333
\(\large \lfloor \log_{10} 60^{12}\rfloor\color{red}{+1}\)
mathmath333
  • mathmath333
2
ganeshie8
  • ganeshie8
how many digits are there in \(\large 10^2\) ?
mathmath333
  • mathmath333
3
mathmath333
  • mathmath333
i see
ganeshie8
  • ganeshie8
how many digits are there in \(\large 10^{300}\) ?
mathmath333
  • mathmath333
301
ganeshie8
  • ganeshie8
how did you get 301 ?
mathmath333
  • mathmath333
add +1
ganeshie8
  • ganeshie8
You got 301 by adding 1 to the logarithm, 300.
ganeshie8
  • ganeshie8
It helps to keep in mind the actual definition of logarithm : logarithm = exponent
ganeshie8
  • ganeshie8
\[\large 10^{\color{red}{300}} \] \(\color{red}{300}\) is called the logarithm of \( 10^{\color{red}{300}}\) relative to base \(10\)
ganeshie8
  • ganeshie8
as you can see, the number of digits in the decimal expansion of \( 10^{\color{red}{300}}\) equals \[\large \lfloor \log_{10} 10^{\color{red}{300}}\rfloor + 1\]
mathmath333
  • mathmath333
that was easy
mathmath333
  • mathmath333
\(log _{10} 10\)
ganeshie8
  • ganeshie8
\(\large 10 = 10^\color{red}{1}\) so the logarithm of \(10\) is \(\color{red}{1}\) (relative to base 10)
mathmath333
  • mathmath333
is it important to include floor function
ganeshie8
  • ganeshie8
how many digits are there in \(\large 3^4\) ?
mathmath333
  • mathmath333
2
ganeshie8
  • ganeshie8
how did u get 2 ?
ganeshie8
  • ganeshie8
http://www.wolframalpha.com/input/?i=floor%28log%2810%2C3%5E4%29%29%2B1
mathmath333
  • mathmath333
81=3^4
ganeshie8
  • ganeshie8
try using the formula
ganeshie8
  • ganeshie8
\[\large \lfloor \log_{10} 3^{4}\rfloor+1\]
ganeshie8
  • ganeshie8
u will see why floor function is necessary
mathmath333
  • mathmath333
ok
ganeshie8
  • ganeshie8
http://math.stackexchange.com/questions/231742/proof-how-many-digits-does-a-number-have-lfloor-log-10-n-rfloor-1
mathmath333
  • mathmath333
it seems i need to remember log values to use the formula
ganeshie8
  • ganeshie8
not worth it, there must be a simpler way to solve it
ganeshie8
  • ganeshie8
Remember, if you're struggling too hard, then it means only one thing : your method is wrong/not clever
mathmath333
  • mathmath333
ok thnx
ganeshie8
  • ganeshie8
is this really supposed to be solved w/o calculator ?
mathmath333
  • mathmath333
yes lol
ganeshie8
  • 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
mathmath333
  • mathmath333
log6=log2+log3
mathmath333
  • mathmath333
i just need to rote log 2 and log 3
ganeshie8
  • ganeshie8
i got better things to memorize than these log values haha
mathmath333
  • mathmath333
its matter of pattern , not rote learning
mathmath333
  • mathmath333
log 2=0.30 , log3=0.47 easy
ganeshie8
  • ganeshie8
Okay, then lets pretend that we have access to log2 and log3 values :)
ganeshie8
  • ganeshie8
log6 = log2 + log3 = 0.30 + 0.47 = 0.77 ?
mathmath333
  • mathmath333
yep correct
ganeshie8
  • ganeshie8
good, plug them in the formula and see..
mathmath333
  • mathmath333
http://www.wolframalpha.com/input/?i=log_%7B10%7D6%3D
ganeshie8
  • ganeshie8
\[\large{\begin{align} \lfloor \log_{10} 60^{12}\rfloor+1 &= \lfloor 12*\log_{10} 60\rfloor+1\\~\\ &= \lfloor 12*(\log_{10}10 +\log_{10} 6)\rfloor+1\\~\\ &\approx \lfloor 12*(1 +0.77)\rfloor+1\\~\\ &=21+1 \end{align}}\]
mathmath333
  • mathmath333
ganeshie8
  • ganeshie8
Ahh you wanto use taylor series is it \[\ln(1+x)=-\sum_{n=1}^\infty\frac{(-x)^n}{n}=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots\]
mathmath333
  • mathmath333
i just will remember values to save time
ganeshie8
  • ganeshie8
do let me know if/when you find a better method
mathmath333
  • mathmath333
yep .
phi
  • 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.

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