How in the world do you do some of these calculations without a calculator?

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I'm using the Kaplan book for general chemistry, and just wondering when there are log functions involved and exponents how the hell do you do some of these calculations without a calculator? I don't see how in the world I can get these answers correct in this section or the physics section without using a calculator especially since there is a time limit.

Thanks.

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Don't focus too much on the hard calculations. Just understand the concept really well. I didn't take the real test yet. But I did many practice tests and I didn't encounter any hard calculation.
 
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Here you go:
http://forums.studentdoctor.net/threads/berkeley-reviews-unique-approach.980939/

the log trick has really helped me:
Log Math Trick
The Know Your Primes Method
First and foremost, you should always look at the answer choices and see how much precision you need. If the answers are far apart, then you can afford to make less rigorous approximations. But if they are close to one another, as they could be from time-to-time, then precision is necessary.

For determining pH from [H+], or any other conversion that involves taking a negative log, we use the following relationship.

- log (a x 10-b) = b - log a

This is applicable for pH, pOH, pKa, and pKb.

Next we teach the know your primes approach. Know the following four logs for approximating a best answer:

log 2 = 0.30
log 3 = 0.48
log 5 = 0.70
log 7 = 0.85

Because prime numbers can be multiplied together to get other numbers, if you need precision you can build from those numbers. And the prime numbers between 1 and 10 will give you the necessary precision to make a good choice on 99.9999999% of the MCAT questions you'll see.

Given Ka = 4.61 x 10-7; pKa = 7 - log 4.61 which is slightly larger than 7 - log 5 = 6.3. So guessing around 6.33 +/- is going to be as much precision as you could need on the MCAT.

Given [OH-] = 2.77 x 10-4; pOH = 4 - log 2.77 which is slightly larger than 4 - log 3 = 3.52 but not as large as 4 - log 2 = 3.7. So guessing around 3.56 +/- is closer than you will likely need.

Given [H+] = 7.93 x 10-3; pH = 3 - log 7.93 which is slightly smaller than 3 - log 7 = 2.15. So guessing around 2.11 +/- is good enough. This is where the proponents of precision will say that knowng 3 - log 8 = 2.10 gets you a more accurate answer. And I can't deny that 2.10 is closer to 2.097 than 2.11, but if the MCAT choices are so close that 2.10 beats 2.11, then the test would have changed so much you would have heard someone complain about log details.

Given Kb = 6.11 x 10-8; pKa = 8 - log 6.11 which is larger than 8 - log 7 = 7.15, but less than 8 - log 5 = 7.3. So guessing around 7.23 +/- is a winning approximation.

Picking the method that works for you is important, because you have to balance the need for speed with your level of satisfaction with an answer before you can move on without lingering second thoughts. The know your primes approach is a great method to find that balance.
 
There are calculation tricks at the end of the Physics book in Kaplan. With logs, it's more of getting it to the right order of magnitude, as often on the MCAT you will encounter answers that differ by more than one order of magnitude and if you know your answer should be 10^15, then it doesn't matter whether it's 1.3*10^15 or 7.3*10^15.
 
I doubt you will ever need to do this in a real exam. If you do, then you are doing something wrong.
 
agreed, and TBR method for log calculations was extremely helpful to me for determining that. I had to waste a ton of time before using TBR method for it.

I'm not really sure what method that is but you should at least be familiar with properties of logs. Such that it becomes immediately obvious that -log(10^-3) = (-3)*(-log(10)) = (-3)*(-1) = 3. There's no "method" required to calculate this save understanding the properties of logs.
 
I'm not really sure what method that is but you should at least be familiar with properties of logs. Such that it becomes immediately obvious that -log(10^-3) = (-3)*(-log(10)) = (-3)*(-1) = 3. There's no "method" required to calculate this save understanding the properties of logs.
right, that's exactly what I posted above. the OP asked for advice on doing log calculations without a calculator.
 
right, that's exactly what I posted above. the OP asked for advice on doing log calculations without a calculator.

No, what you posted was an in-depth method estimating just about any log. It's a great method and all, but chances are, OP will not need those in depth. As another poster above said, if you're trying to do really complex calculations, you did something wrong.
 
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