Logarithm question...

[tncekm]

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

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[Chemist0157]

Yep.

Basically,

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

 

[fileserver]

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)

 

[iA-MD2013]

looks good

 

[tncekm]

Awesome, thx everyone.

 

[werd]

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]

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!