# Math Help - Check my work

1. ## Check my work

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

2. ## note

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

3. 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

4. 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)

5. 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

6. 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.

7. 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

8. 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.

9. 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

10. 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...

11. 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

12. 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...