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mathmath333
 one year ago
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mathmath333
 one year ago
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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \color{black}{\begin{align} \normalsize \text{Find the number of digits in }\ 60^{12}\hspace{.33em}\\~\\ \end{align}}\)

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1438031157075:dw

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1but calclulting whole stuff is cumbersome

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3dumb q, but are you allowed to use calculator for taking logarithm

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i mean my question of asking whether you're allowed to use a calculator is dumb :)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i mean no cheating stuff allowed

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3number 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
 one year ago
Best ResponseYou've already chosen the best response.1where did u get this formula

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1what is the name of the formula

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Its not a formula as such, just work few examples till you get convinced, keeping in mind the definition of logarithm \[\large 10^x = N ~~\iff~~ x=\log_{10} N\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how many digits are there in \(\large 10^1\) ?

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1\(\large \lfloor \log_{10} 60^{12}\rfloor\color{red}{+1}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how many digits are there in \(\large 10^2\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how many digits are there in \(\large 10^{300}\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how did you get 301 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3You got 301 by adding 1 to the logarithm, 300.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3It helps to keep in mind the actual definition of logarithm : logarithm = exponent

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\large 10^{\color{red}{300}} \] \(\color{red}{300}\) is called the logarithm of \( 10^{\color{red}{300}}\) relative to base \(10\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3as you can see, the number of digits in the decimal expansion of \( 10^{\color{red}{300}}\) equals \[\large \lfloor \log_{10} 10^{\color{red}{300}}\rfloor + 1\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(\large 10 = 10^\color{red}{1}\) so the logarithm of \(10\) is \(\color{red}{1}\) (relative to base 10)

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1is it important to include floor function

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3how many digits are there in \(\large 3^4\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3http://www.wolframalpha.com/input/?i=floor%28log%2810%2C3%5E4%29%29%2B1

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3try using the formula

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\large \lfloor \log_{10} 3^{4}\rfloor+1\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3u will see why floor function is necessary

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1it seems i need to remember log values to use the formula

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3not worth it, there must be a simpler way to solve it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Remember, if you're struggling too hard, then it means only one thing : your method is wrong/not clever

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3is this really supposed to be solved w/o calculator ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i don't really see how one computes it without knowing the value of \(\log_{10}6\), or without working the actual number hmm

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1i just need to rote log 2 and log 3

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3i got better things to memorize than these log values haha

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1its matter of pattern , not rote learning

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1log 2=0.30 , log3=0.47 easy

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Okay, then lets pretend that we have access to log2 and log3 values :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3log6 = log2 + log3 = 0.30 + 0.47 = 0.77 ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3good, plug them in the formula and see..

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\[\large{\begin{align} \lfloor \log_{10} 60^{12}\rfloor+1 &= \lfloor 12*\log_{10} 60\rfloor+1\\~\\ &= \lfloor 12*(\log_{10}10 +\log_{10} 6)\rfloor+1\\~\\ &\approx \lfloor 12*(1 +0.77)\rfloor+1\\~\\ &=21+1 \end{align}}\]

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.1http://openstudy.com/users/mathmath333#/updates/55b3954fe4b071e6530e16e9 one more method for finding log

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Ahh 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
 one year ago
Best ResponseYou've already chosen the best response.1i just will remember values to save time

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3do let me know if/when you find a better method

phi
 one year ago
Best ResponseYou've already chosen the best response.1you 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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