View Full Version : Logarithm question...


tncekm
03-06-2008, 09:35 PM
Okay, so I'm trying to derive the Hendersen Hasselbach equation, and its been a LONG time since I've worked with logarithms, so I want to make sure I'm doing this right.

I couldn't find an explicit agreement with this via google, so maybe someone here can verify that I've been doing this right (it gives me the right derivation)

Log(A/B) = -Log(B/A) b/c

Log(A/B) = LogA - LogB -->
-Log(A/B) = LogB - LogA -->
-Log(A/B) = Log(B/A) -->
Log(A/B) = -Log(B/A), correct?

Thx

EDIT: Just realized I constructed the equation wrong, so I didn't even need this to get HH eqn. But, I'm still interested in whether or not this is a valid proof :D

Chemist0157
03-06-2008, 10:02 PM
Yep.

Basically,

LOG(1/X) = -LOG(X)

fileserver
03-06-2008, 10:29 PM
yes, that is correct.

or you can do this:

" Log(A/B) = -Log(B/A) b/c "

= - log (b/a)
= - log (b^1 a^-1)
= log (b^1 a^-1)^-1
= log (b^-1 a^1)
= log (a/b)

jgberken
03-06-2008, 11:16 PM
looks good :)

tncekm
03-06-2008, 11:16 PM
Awesome, thx everyone.

werd
03-07-2008, 06:56 PM
it's always bugged me that hendersen/hasslebach got this 'famous' equation named after them... all they did was rearrange the equillibrium equation. it wasn't even a complex rearrangement; it can be empirically derived in about 45 seconds using algebra. oh well.

tncekm
03-08-2008, 01:35 AM
it's always bugged me that hendersen/hasslebach got this 'famous' equation named after them... all they did was rearrange the equillibrium equation. it wasn't even a complex rearrangement; it can be empirically derived in about 45 seconds using algebra. oh well.
Yeah... I agree.

If you want to talk about WEAK how about the rest mass energy equation: E=mc² Its only three letters! :D