doing LOGS in your head?

[tayloreve]

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Hey guys,

I just watched Chad's genchem video on pH & pOH.

When he asks you to calculate: pH = -log(8.6 x 10^-6), Chad very easily does the -log of this number in his head by approximating. However, his method really confuses me and with the pressure of the DAT and the time constraints, I don't think doing it his way is a choice for me.

What tricks do you guys know to be able to calculate logs in your head? How do you do it!?

I really wish we could use a calculator on this part, blah -_-.

Starting to get a little discouraged, thanks all.

 

[ktran17]

I memorized log 1 to 10 and use those values to estimate a log problem
I did the same thing for squareroot 1-10 as well

 

[tayloreve]

I memorized log 1 to 10 and use those values to estimate a log problem
I did the same thing for squareroot 1-10 as well

so did you basically do it how Chad taught it?

 

[ScarletKnight24]

Hey guys,

I just watched Chad's genchem video on pH & pOH.

When he asks you to calculate: pH = -log(8.6 x 10^-6), Chad very easily does the -log of this number in his head by approximating. However, his method really confuses me and with the pressure of the DAT and the time constraints, I don't think doing it his way is a choice for me.

What tricks do you guys know to be able to calculate logs in your head? How do you do it!?

I really wish we could use a calculator on this part, blah -_-.

Starting to get a little discouraged, thanks all.

I was having a hard time when I first started but now I can calculate everything in my head very quickly.

lets use his example
[H+] = 8.6 x 10^-6

now that value is very close to 10 x 10^-6 OR 1.0 x 10^-5

since the ph of 1.0 * 10^-5 is 5 and your concentration is a bit less than 1.0 * 10^-5 that means your pH would be a little bit above 5

it is an inverse relationship since ph = -log [H+]


Another example:
let say I have a concentration that is 2.5 * 10^-3
now 2.5 * 10^-3 is pretty close to 1 * 10^-3 and since the ph of 1*10^-3 is 3 and 2.5 *10^-3 is a bit higher than that so my ph would be a bit lower than 3.


I was going crazy when I first started too but you'll get the hang of it pretty quick. Just practice.

NO need to memorize the log table.

 

[n27s]

i struggled with this also, until i figured out a way to do it that's so easy, you could be brain dead and do it

so if the number is 8.6 x 10^-6, I focus on the 10^-6, I put a 5 on top if it (in my head i picture a column of numbers going down 1, 2, 3, 4..) so 5 would be on top of 6, now after 5 comes 6, i put the 6 under the number. so on your paper you should have the number written down with a 5 on top of the 10^-6 and 6 written underneath it. This tells you your -log will be between 5 and 6. The how much more part is easy, since it's 8/10, think inverse...5.2 or 5.1 I would guess. Works every time.

 

[Muggs]

i struggled with this also, until i figured out a way to do it that's so easy, you could be brain dead and do it

so if the number is 8.6 x 10^-6, I focus on the 10^-6, I put a 5 on top if it (in my head i picture a column of numbers going down 1, 2, 3, 4..) so 5 would be on top of 6, now after 5 comes 6, i put the 6 under the number. so on your paper you should have the number written down with a 5 on top of the 10^-6 and 6 written underneath it. This tells you your -log will be between 5 and 6. The how much more part is easy, since it's 8/10, think inverse...5.2 or 5.1 I would guess. Works every time.

I use a similar method. Just to add to this, I first look at the exponent of the number (-6 in this case). All you have to do is add 1 to this number, and your final pH will be between these two numbers: 5 and 6 (ignore the negative sign). Always remember then that a pH of 6 is the -log(1x10^-6). Due to the inverse relationship between [H+] and pH, since you have an [H+] higher than 1x10^-6 (8.6x10^-6 in this case), that will mean that the actual pH will come out much closer to the lower part of your estimated pH range: closer to pH 5. If the [H+] was 2.0x10^-6, it's not that much more than 1x10^-6, so the pH will be closer to the higher part of the estimated pH range: pH 6.

Hope this helps.

 

[ktran17]

so did you basically do it how Chad taught it?

not exactly. more like scarlett's method.
I use this method
http://w w w .curiousmath. c o m/index.php?name=News&file=article&sid=32

 

[denticle]

i struggled with this also, until i figured out a way to do it that's so easy, you could be brain dead and do it

so if the number is 8.6 x 10^-6, I focus on the 10^-6, I put a 5 on top if it (in my head i picture a column of numbers going down 1, 2, 3, 4..) so 5 would be on top of 6, now after 5 comes 6, i put the 6 under the number. so on your paper you should have the number written down with a 5 on top of the 10^-6 and 6 written underneath it. This tells you your -log will be between 5 and 6. The how much more part is easy, since it's 8/10, think inverse...5.2 or 5.1 I would guess. Works every time.

what do you mean by inverse of 8/10?

 

[n27s]

what do you mean by inverse of 8/10?

i mean 8.... is on the large side on a number scale of 1 to 10 so the decimal value will be the inverse of a large number.. aka a small number.. (I'm refering the decimal place that comes after 5 in the ph) .. around 5.1 or 5.2the ph wont be 5.8 or something closer to 6. sorry if i dont make sense. my brain is fried right now.

in other words if it was something like 2.1 x 10^-6 the ph would be closer to 6 (bc 2 is a small value, the decimal value will be larger) 5.7 or 5.8, if you're off by a one or two decimal spaces it doesn't really effect picking out the right answer

 

[FancyFloss]

Bump

 

[Likkriue]

The easiest method and least confusing is to know the magic number 3.16.

3.16 X 10 to "X" power will always be (X-1).5.
For example 3.16 X 10 ^-7 will be 6.5.
And 3.16 X 10 ^ -5 will be 4.5.

So let's say you have 1 X 10 ^ -5, you know that's 5. But if you had 9 X 10^ -5, it would have to be 4.1, why? Because the midpoint 3.16 X 10^ -5 will always be 4.5. So now you realize the higher the value the lower the ph. Cause the H+ concentration goes up.

 

[BiigNerd]

You guys. I found the holy grail of log approximations. Search for "logarithm simplification for everyone!" in the MCAT Discussion by PBandJ. His method only requires only memorizing log(2) and log(3), then from there you just use the properties of logs (you should know them already).