I've enclosed an excel program I made to find all different possible human dna.
I would like someone to check the work please.
note: you need to click on the green tab to get to what I want checked
I'm not asking you to check my data, what I am asking is if...
[ ( 2 ^ r ) ^ g ] ^ c
[ 2 ^ ( r • g ) ] ^ c
2 ^ ( r • g • c )
2 ^ ( 5000 • 543.4782609 • 46 )
2 ^ ( 2717391.304 • 46 )
2 ^ 125000000
( 2 ^ 1000 ) ^ 125000
( 1.0715 • 10 ^ 301 ) ^ 125000
1.0715 ^ 125000 • 10 ^ ( 301 • 125000 )
( 1.0715 ^ 1000 ) ^ 125 • 10 ^ 37625000
( 9.82144 • 10 ^ 29 ) ^ 125 • 10 ^ 37625000
9.82144 ^ 125 • 10 ^ ( 29 • 125 ) • 10 ^ 37625000
9.82144 ^ 125 • 10 ^ 3625 • 10 ^ 37625000
9.82144 ^ 125 • 10 ^ ( 3625 + 37625000 )
9.82144 ^ 125 • 10 ^ 37628625
( 9.82144 ^ 100 ) ^ 1.25 • 10 ^ 37628625
( 1.6501 • 10 ^ 99 ) ^ 1.25 • 10 ^ 37628625
1.6501 ^ 1.25 • 10 ^ ( 99 • 1.25 ) • 10 ^ 37628625
1.870199043 • 10 ^ 123.75 • 10 ^ 37628625
1.870199043 • 10 ^ ( 123.75 + 37628625 )
1.870199043 • 10 ^ 37628748.75
is correct.
in other words...
does $\displaystyle 2^{125000000}=1.870199043 \cdot 10^{37628748.75}$
In otherwords is:Originally Posted by Quick
$\displaystyle 125000000=\log_2(1.870199043)+37628748.75\times \log_2(10)$?
Is that close enough? All the arithmetic is double precision and as allCode:> >logbase(1.870199043,2)+logbase(10,2)*37628748.75 124999998.551 > >
quantities involved are positive I expect this is good to at least 10 significant
digits.
Checking this in an arbitrary precisions arithmetic package tell me all the
quoted figures above are correct (if I have typed the numerical constants
correctly)
RonL
How do you use logarithims, it seem to me that if you have something like 2^5=... than you would get rid of the two and multiply everything by log_2 and then you can use the rules of logarithims to say that log(x^y)=y*log(x), am I right? (I'm not really sure about the rules of Logarithims)Originally Posted by CaptainBlack
if $\displaystyle 2^x=y$, then $\displaystyle x=\log_2(y)$,Originally Posted by Quick
and for an arbitary base:
if $\displaystyle b^x=y$, then $\displaystyle x=\log_b(y)$.
Basic rules of logarithms can be derived from the rules for exponents:
$\displaystyle \log_b(x.y)=\log_b(x)+\log_b(y)$,
$\displaystyle \log_b(x^n)=n\log_b(x)$,
$\displaystyle \log_a(x)=\frac{\log_b(x)}{\log_b(a)}$
RonL
Another way to think of logarithms is of functions.
The function $\displaystyle y=a^x$ $\displaystyle a>0$ has an inverse function (I assume you know what that means).
You define that function as,
$\displaystyle y=\log_a x$
-----
An inverse function is one such as,
$\displaystyle f(f^{-1}(x))=f^{-1}(f(x))=x$
The interesting property between exponent and logarithms is that they express a sum as a product and a product as a sum respectively.
shouldn't it be $\displaystyle \log_a(y)=x$Originally Posted by ThePerfectHacker
you assume too much! I have been trying to figure what an inverse function is ever since I got here, I've been starting to believe that the inverse of a function is a derivative of the function, which I would like to know more about...I assume you know what that means
Originally Posted by CaptainBlack
Give a break to a high-school freshman.Originally Posted by CaptainBlack
NoOriginally Posted by Quick
When you come back I can give you an online lecture of what an inverse function is. When I was younger I also never understood it (because high schools is too informal). It makes perfect sense with setsOriginally Posted by Quick