antilogs (quick way) ?

[SaintJude]

Hi,

Does anyone have a quick way to calculate anti-logs.

Say you're given a pKa of a weak acid is 5.7 and you need to then take the antilog of pKA to get the Ka. Any trick to do this quickly?

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[milski]

The fraction part gives you a number between 1 and 10. It may be useful to memorize a few numbers like 0.3 -> 2, 0.7 -> 5 and 0.9 -> 8. The just multiply that number by 10 to the integer part of the log.

For example, anti log of 2.7 is 5*100=500.

One thing to keep in mind that anti log is rapidly increasing, which means that you have a value roughly in the middle between other two, its antilog will be closer to the antilog of the smaller one. In other words antilog of 0.8 is closer to 5 then to 8. Same way, antilog of 0.5 is closer to 1 than to 10.

 

[MedPR]

Hi,

Does anyone have a quick way to calculate anti-logs.

Say you're given a pKa of a weak acid is 5.7 and you need to then take the antilog of pKA to get the Ka. Any trick to do this quickly?

Would you ever really need to know Ka more specifically than 1*10^-5.7?

It's bigger than 1*10^-6, but smaller than 1*10^-5. I don't know anymore than that without going to a calculator though.

 

[mcloaf]

Would you ever really need to know Ka more specifically than 1*10^-5.7?

It's bigger than 1*10^-6, but smaller than 1*10^-5. I don't know anymore than that without going to a calculator though.

I'm with you on this one.

 

[SaintJude]

Yeah, I guess y'all are right. I estimated the value as ~ 5 x 10^-6 rather than just 1 x 10^-6. Because sometimes the differences b/w values of answer choices (at least in Kaplan) are quite close. I think I was therefore referring to milski's values. Thanks !

log 2= 0.3
log 5= 0.7
log 8=0.9

 

[milski]

Yeah, I guess y'all are right. I estimated the value as ~ 5 x 10^-6 rather than just 1 x 10^-6. Because sometimes the differences b/w values of answer choices (at least in Kaplan) are quite close. I think I was therefore referring to milski's values. Thanks !

log 2= 0.3
log 5= 0.7
log 8=0.9

To be honest, the only one I "know" is log 3 = 0.5 from sqrt 10 = 3. I could come up with good estimates for 0.3 and 0.25/0.75 on the same principle. The rest can be interpolated from there. I'll be surprised you need that much precision for the test. But I have not taken it, so don't take my word for granted.

 

[MedPR]

Yeah, I guess y'all are right. I estimated the value as ~ 5 x 10^-6 rather than just 1 x 10^-6. Because sometimes the differences b/w values of answer choices (at least in Kaplan) are quite close. I think I was therefore referring to milski's values. Thanks !

log 2= 0.3
log 5= 0.7
log 8=0.9

Yea those are in TBR too, along with log3 = .47

I guess I don't understand how antilog works. How would you use known log values to estimate that 10^-5.7 is 1.99*10^-6?

 

[milski]

Yea those are in TBR too, along with log3 = .47

I guess I don't understand how antilog works. How would you use known log values to estimate that 10^-5.7 is 1.99*10^-6?

10^(-5.7)=10^(-6+0.3)=10^-6 * 10^0.3=10^-6*2

Be careful when it's antilog from a negative number, since you want the fraction to be positive in that case. For our example you want -5.7=-6+0.3 and not -5.7=-5-0.7

While the second is correct, it would make you calculate 10^-5/10^0.7 which is not the quick way we were going for.

 

[MedPR]

10^(-5.7)=10^(-6+0.3)=10^-6 * 10^0.3=10^-6*2

Be careful when it's antilog from a negative number, since you want the fraction to be positive in that case. For our example you want -5.7=-6+0.3 and not -5.7=-5-0.7

While the second is correct, it would make you calculate 10^-5/10^0.7 which is not the quick way we were going for.

Ok so 10^0.3 * 10^-6 = 2*10^-6, which is the same thing as 2*log(-6)?

 

[milski]

Ok so 10^0.3 * 10^-6 = 2*10^-6, which is the same thing as 2*log(-6)?

Sort of:
10^0.3 * 10^-6 = 2 * 10^-6
log(10^0.3) + log(10^-6) = log(2) + log(10^-6)
0.3 + -6 = log(2) + -6


log(-6) is undefined

 

[MedPR]

Sort of:
10^0.3 * 10^-6 = 2 * 10^-6
log(10^0.3) + log(10^-6) = log(2) + log(10^-6)
0.3 + -6 = log(2) + -6


log(-6) is undefined

I meant 2*log(10^-6), but that's wrong anyway.

10*^0.3 * 10^-6 = log(10^0.3) + log(10^-6) = log(2) + log(10^-6) = .3 + -6 = -5.7.

Thanks!

So when you have something like log6, that's the same thing as log3 + log2.

And when you have 10^x * 10^y, that's the same thing as logx + logy also

 

[milski]

I meant 2*log(10^-6), but that's wrong anyway.

10*^0.3 * 10^-6 = log(10^0.3) + log(10^-6) = log(2) + log(10^-6) = .3 + -6 = -5.7.

Thanks!

So when you have something like log6, that's the same thing as log3 + log2.

And when you have 10^x * 10^y, that's the same thing as logx + logy also

Yes on the first line, sort of on the second. You probably get it but are skipping when writing it here.

10^x * 10^y is just that. log(10^x * 10^y) = log(10^x)+log(10^y)=x+y