question on approximating logs
Started by spoog74
log 1.0 x 10^3 is also 3...the way I do it is...bring the 3 from 10^3 down and add with log 1.0...and log 1 is 0
so basically for any log a x b^n = n + log a...now log a should be between 0 and 1..
so for example log 3.5 x 10^-58 = -58 + log 3.5, so the final answer should be between -57 and -58..
Now if you put a negative sign in the beginning of the log, the sign of the equation changes, - log 3.5 x 10^-58 = 58 - log 3.5 and the answer will be between 57 and 58..
pertaining to what i was asking;
for a question such as this one;
What is the pH of a solution that contains 0.003M HF and 0.02M NaF (Ka, HF=6.8x10-4)?
i know we need to figure out the pka which i rounded to be around 3.2 then i get the rest as
ph = 3.2 + log [ .02] / [.003]
How would you get the log of those 2 numbers quick in your head?
Dude..just curious, why are you doing it this way...isnt this a Common Ion effect problem...
Regarding your question, we will need a 3-4 Phd in Logs to calculate that ratio in our head, the best way will be to look at the answer choice..
From the log rules, Log a/b = log a - log b....so the ratio of log 0.02 - log 0.003 should not be too much higher than 3.2 (your calculate Pka)...is there a answer choice close to 4 or something..
Regarding your question, we will need a 3-4 Phd in Logs to calculate that ratio in our head, the best way will be to look at the answer choice..
From the log rules, Log a/b = log a - log b....so the ratio of log 0.02 - log 0.003 should not be too much higher than 3.2 (your calculate Pka)...is there a answer choice close to 4 or something..
Hey all-
First post to SDN, prepping for DAT July 7. To the OP: here's a little trick I just picked up from my online Kaplan class the other day.
To do a quick calc of -log of any #...
-log [ n x 10^-m ] = (m-1).(10-n)
For example, to get the pH of an acid whose [H+] is [3.6x10^-7]...
pH = -log [3.6x10^-7] = (7-1).(10-3.6) = 6.64 (actual calculator value is 6.44 i think)
Hope that helps!
First post to SDN, prepping for DAT July 7. To the OP: here's a little trick I just picked up from my online Kaplan class the other day.
To do a quick calc of -log of any #...
-log [ n x 10^-m ] = (m-1).(10-n)
For example, to get the pH of an acid whose [H+] is [3.6x10^-7]...
pH = -log [3.6x10^-7] = (7-1).(10-3.6) = 6.64 (actual calculator value is 6.44 i think)
Hope that helps!
i know i'm bringing this thread back but does this last method work? i'm having some trouble approximating logs as well
thanks
thanks
How to calculate logs in your head in less than a second.
For pH calculations and what not I just think of the exponent and the base number being inversely proportional.
For example: 4.2x10^-8 The higher the base number is (4.2) the lower the exponent (8) will be. So i'd estimate this one to be 7.6. Lets see what happens when I make the base number even higher.
Another example: 5.2x10^-8 Higher base number (5.2), lower the exponent (8). This case probably ~7.5 ish. So notice as I raise the base number, the exponent value gets smaller?
This isn't mathematically correct because -8 to -7 is getting larger, it just enables me to do the calculation as fast as subtracting two single digit numbers in a quick glance. Hope that helps.
For pH calculations and what not I just think of the exponent and the base number being inversely proportional.
For example: 4.2x10^-8 The higher the base number is (4.2) the lower the exponent (8) will be. So i'd estimate this one to be 7.6. Lets see what happens when I make the base number even higher.
Another example: 5.2x10^-8 Higher base number (5.2), lower the exponent (8). This case probably ~7.5 ish. So notice as I raise the base number, the exponent value gets smaller?
This isn't mathematically correct because -8 to -7 is getting larger, it just enables me to do the calculation as fast as subtracting two single digit numbers in a quick glance. Hope that helps.
