Check my work

**Quick** · July 19th 2006, 07:55 PM

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

**Quick** · July 20th 2006, 04:41 PM

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 $2^{125000000}=1.870199043 \cdot 10^{37628748.75}$

**CaptainBlack** · July 21st 2006, 09:33 PM

Originally Posted by Quick

in other words...
does $2^{125000000}=1.870199043 \cdot 10^{37628748.75}$

In otherwords is:

$125000000=\log_2(1.870199043)+37628748.75\times \log_2(10)$ ?

Code:

>
>logbase(1.870199043,2)+logbase(10,2)*37628748.75
      124999998.551 
>
>

Is that close enough? All the arithmetic is double precision and as all
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

**Quick** · July 22nd 2006, 05:11 AM

Originally Posted by CaptainBlack

In otherwords is:

$125000000=\log_2(1.870199043)+37628748.75*\log_2(1 0)$ ?

Code:

>
>logbase(1.870199043,2)+logbase(10,2)*37628748.75
      124999998.551 
>
>

Is that close enough? All the arithmetic is double precision and as all
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)

**CaptainBlack** · July 22nd 2006, 09:51 AM

Originally Posted by Quick

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)

if $2^x=y$ , then $x=\log_2(y)$ ,

and for an arbitary base:

if $b^x=y$ , then $x=\log_b(y)$ .

Basic rules of logarithms can be derived from the rules for exponents:

$\log_b(x.y)=\log_b(x)+\log_b(y)$ ,

$\log_b(x^n)=n\log_b(x)$ ,

$\log_a(x)=\frac{\log_b(x)}{\log_b(a)}$

RonL

**ThePerfectHacker** · July 22nd 2006, 08:01 PM

Another way to think of logarithms is of functions.
The function $y=a^x$ $a>0$ has an inverse function (I assume you know what that means).
You define that function as,
$y=\log_a x$
-----
An inverse function is one such as,
$f(f^{-1}(x))=f^{-1}(f(x))=x$

The interesting property between exponent and logarithms is that they express a sum as a product and a product as a sum respectively.

**CaptainBlack** · July 22nd 2006, 08:03 PM

Originally Posted by ThePerfectHacker

Another way to think of logarithms is of functions.
The function $y=a^x$ $a>0$ has an inverse function (I assume you know what that means).
You define that function as,
$y=\log_a x$
-----
An inverse function is one such as,
$f(f^{-1}(x))=f^{-1}(f(x))=x$

The interesting property between exponent and logarithms is that they express a sum as a product and a product as a sum respectively.

What I find surprising is that Quick, who displays significant mathematical
knowlege should be lacking in the logarithm department.

RonL

**Quick** · July 23rd 2006, 12:48 AM

Originally Posted by ThePerfectHacker

Another way to think of logarithms is of functions.
The function $y=a^x$ $a>0$ has an inverse function (I assume you know what that means).
You define that function as,
$y=\log_a x$

shouldn't it be $\log_a(y)=x$

I assume you know what that means

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...

Originally Posted by CaptainBlack

What I find surprising is that Quick, who displays significant mathematical
knowlege

Originally Posted by CaptainBlack

should be lacking in the logarithm department.

Give a break to a high-school freshman.

**ThePerfectHacker** · July 23rd 2006, 08:07 AM

Originally Posted by Quick

shouldn't it be $\log_a(y)=x$

No

Originally Posted by Quick

what an inverse function is ever since I got here,

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 sets

**Quick** · July 30th 2006, 06:30 PM

just a quick question: does $\log_bb=1$ ? (that's the impression I'm getting but I want to be sure)

Originally Posted by ThePerfectHacker

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 sets

I'm back...

**ThePerfectHacker** · July 30th 2006, 06:52 PM

Originally Posted by Quick

just a quick question: does $\log_bb=1$ ? (that's the impression I'm getting but I want to be sure)

Yes

**Quick** · July 31st 2006, 01:54 PM

Originally Posted by ThePerfectHacker

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 sets

I already know what an inverse function is, (there are many websites besides MHF) but none of the sites I've been two explain how to find it...

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