Natural Log..

[Ranelar]

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Dumb question maybe, but I forget how to calculate natural log without a calculator... I ran across it in this problem:

H3AsO3 has a ka of 6.6x10^-10 at room temp, find Gibbs free energy. I did:

G = -RT(lnK)
G = -8(300)[ln(0.00000000066)] rounding for R and T to make life easier.

now I'm stuck without a calculator. I can make both sides an exponent of e, but 2.7^-2400 doesn't seem easy to calculate either 😛 What am I supposed to do?

 

[Streetwolf]

Dumb question maybe, but I forget how to calculate natural log without a calculator... I ran across it in this problem:

H3AsO3 has a ka of 6.6x10^-10 at room temp, find Gibbs free energy. I did:

G = -RT(lnK)
G = -8(300)[ln(0.00000000066)] rounding for R and T to make life easier.

now I'm stuck without a calculator. I can make both sides an exponent of e, but 2.7^-2400 doesn't seem easy to calculate either 😛 What am I supposed to do?

-2400 ln(6.6 * 10^-10)
-2400 [ ln(6.6) - 10*ln(10) ] <-- properties of logs

So now you have to deal with ln(6.6) and ln(10) and you can probably estimate them if you needed to.

Remember you're dealing with 2.71 so that ln(2.71) = 1 and ln(7.3) ~ 2. So ln(6.6) is probably somewhere near 1.85 (actual answer ~1.887).

And since 2.71^3 is approximately 20, you have ln(20) ~ 3. So ln(10) is probably near 2.3 (actual answer ~2.303).

Yeah you have to work with the numbers a bit but at least you can come up with a pretty reasonable guess. They won't ask stuff like that on the test.

And also remember that ln(10) can be turned into base 10 by doing log(10) / log(e) which equals 1/log(e). Unfortunately you have to solve log(e).

That's the law of bases or something like that. I forget the name. Just look it up, you'll find it.

 

[Ranelar]

Thanks, I haven't seen the law used for scientific notation like that before, but I guess it makes sense, very cool 😛