Approximating logs in your head... Really?

[swindoll]

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I'm having trouble learning this. Chad's method didn't help.
How frequently do questions like this pop up on DAT? Thanks!

 

[dhk2131]

It doesn't, but I'd still memorize the logs of 2-9 just in case (it's simple, log 2 = 0.3 (1 more than 2), then log 3-7 is 0.5, 0.6, 0.7, 0.8, 0.85, then 8 and 9 are just 0.05 apart (0.90 and 0.95).

 

[Dr.Penny]

I had the same problem...very frustrating. You really just need to memorize a set of numbers and practice. I took the DAT today and I literally got maybe 1-2 questions...but its different for everyone. Here is something that helped me with anti-logs.

First I memorized this list, which isn't too terrible
log1= 0.0
log2= 0.30
log3= 0.48
log4= 0.60
log 5= 0.70
log 6= 0.78
log 7= 0.85
log 8= 0.90
log 9= 0.95
log 10= 1.00

So this helped me a lot with figuring out ph, poh, etc
say you have to figure out ph= -log (4 x 10^-6)

1) take the opposite sign of -6
2) subtract the log of 4...0.60
3) You get 6-0.60=5.4

If this confuses you dont bother with it...but it was easier for me.

 

[chemkay]

There is a super easy way to do logs (it's how chad showed)
For example: 6.5x10-5

Look at -5 and start with right between that and the lower number, which is 4. So 4.5. If the number is greater than 3 (6.5 in this situation), then go below 4.5. If less than 3, go above 4.5. I would estimate this log to be around 4.25-4.35.

 

[hareman]

Look up the "(n-1).(10-m)" method. It's easier than chads and you'll thank me lol

 

[lianalily]

Look up the "(n-1).(10-m)" method. It's easier than chads and you'll thank me lol

where can I find that method? I tried googling it but no luck

 

[swindoll]

Look up the "(n-1).(10-m)" method. It's easier than chads and you'll thank me lol

I cannot find anything on it online. Could you please explain it a little bit? Thanks in advance!