# Calculating Logs for GChem (from Destroyer)

Discussion in 'DAT Discussions' started by unketa, Jan 4, 2009.

1. ### unketa

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So in destroyer there are a bunch of GChem questions requiring approximation of logs, I "kinda get it" but I think it's mostly a fluke that I my guess-timate is "kinda right." Is there an effective way of doing it?

Take a look at this: -log 2.2 x 10 ^ -7 = 6.65

I understand that since its a -7, the pH will fall in the 6 range. What I don't understand is how to estimate that its .65?

Similarly in these cases:
1. -log 8.6 x 10 ^ -3 = 2.1
2. -log 5 x 10^ -5 = 4.3
3. -log 4.1 x 10^ -5 = 4.1

Hopefully, a GChem Pro can answer my question =)

Thank you!

2. ### doc3232 7+ Year Member

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I can tell you that it will be very obvious on the test. You can round it on the test and still the correct answer will be distinct from the other answer choices.
Don't hassle on the math.

3. ### Streetwolf Ultra Senior Member Dentist 7+ Year Member

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Better yet, a math pro

You should hopefully know (or learn now) that log(AB) = log(A) + log(B). With that in mind, -log(8.6 x 10^-3) = -(log(8.6) + log(10^-3)). Distribute the negative sign and get -log(8.6) - log(10^-3). The term on the right is -3 so it becomes -log(8.6) - (-3) = 3 - log(8.6).

In general if you have -log(x * 10^-y), you'll end up with y - log(x), where x is between 1 and 10 (not including those). If x is not between 1 and 10 then you need to adjust the value in scientific notation to make it between 1 and 10.

Anyhow, you know it is between 2 and 3. Which one is it closer to? I base it on the half. The square root of 10 is 3.16. That means 10^(1/2) = 3.16. So log(3.16) = 1/2. Anything less than 3.16 will be under 1/2, and anything over 3.16 will be above 1/2. So here you have 8.6, which is well above 3.16. That means you are well above 1/2, which means you are subtracting a number greater than 1/2 from 3. So you will end up with something between 2 and 2.5. In 99.9% of cases this should be enough to get your answer. I would say 100 but who knows.

Since you're so close to 10, I'd say you'd end up with about 2.1, which is about what you get.

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4. ### yankees27th 7+ Year Member

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You're not going to have to be that exact. Let's say you have:
-log 8.6 x 10 ^ -3
Forget the 8.6 part. Pretend it's a 1. You get your answer to be 3. Now, I'm assuming this is a pH (or pOH) problem. Now, if your [H+] is more than 1x10^-3 (which it is), you have to move your pH in an acidic direction, so your answer will be less than 3. 8.6 is close to 10 so you move it a lot and you would estimate about 2.1 (I probably would have guessed 2.2 or 2.3 actually...just goes to show you that this isn't so precise). If it was 1.2x10^-3, your answer might be closer to 2.9.

Hopefully that should help. Good luck and let me know if you want me to re-phrase something.

5. ### IdahoDoc 7+ Year Member

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I hope this helps this is how i do -log in my head like you are doing.
You got the first part correct if it is a negative exponent just do one less then the exponent. then for the estimating part if it is a 1.0 X10-.... then it will be one less then the exponent exactly if it is 2.0 then it will be one less then the exponent and .69 here is a quick summary

1.0 = 1
2.0 = .69
3.0 = .52
4.0 = .39
5.0 = .3
6.0 = .22
7.0 = .15
8.0 = .09
9.0 = .04

so example... 5.2X10-8 ~ 7.32
7.8X10-2 ~ 1.11

Hope this helps if you have questions pm me.

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6. OP

### unketa

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Thank you everyone ! This is really helpful !
Much SDN Love =)

7. ### Sublimation 5+ Year Member

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Genius lol. I got so depressed last year when i found out i had to do Logs without a calculator on the DAT. and due to my laziness i didnt put the effort to find out and when i did i stumbled on some really bizarre and incomprehensible methods. But this.....this is truly worthy. THANKS.

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