question on approximating logs

[spoog74]

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i know that -log 1.0x10^-3 would be 3 but how would you do just the log 1.0 x 10^3 ? How would you approximate that without a calculator?

 

[casper22]

i know that -log 1.0x10^-3 would be 3 but how would you do just the log 1.0 x 10^3 ? How would you approximate that without a calculator?

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..

 

[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..

Perfect. Ty

 

[casper22]

Perfect. Ty

 

[spoog74]

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?

 

[casper22]

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..

 

[casper22]

Dude..just curious, why are you doing it this way...isnt this a Common Ion effect problem...

N/M I didnt read the question correctly...my bad

 

[tommyinVA]

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!

 

[torobcheh21]

i know i'm bringing this thread back but does this last method work? i'm having some trouble approximating logs as well

thanks

 

[Josh779]

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.