Im working through the Destroyer books and in QR and GC I'm struggling on the problems where it's required to know the answer to a more complex logarithmic function. For example it was expected I know that -log 3 x 10^-7 is approximately 6.5; or -log 9.5 x 10^-4 ~ 3. I do not know or really understand a way to estimate these more complex log functions. For the DAT do I need to know these? Or am I okay knowing just simple -log 1 x 10 ^-5 = 5 sort of things. If I do need to know these, can someone please explain how to go about learning/memorizing them?
Knowing answers to more complex log functions
- Thread starter marisam
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You don't need to memorize them, there's a simple way to do it. You already know that -log 1 x 10^-1 = 1, -log 1 x 10^-2 = 2, -log 1 x 10^-3 = 3, etc.. So using this logic, you can estimate what -log 3 x 10^-7 is by saying that -log 3 x 10^-7 is in between -log 1 x 10^-7 and -log 10 x 10^-7 (because 3 is in between 1 and 10). Then, convert -log 10 x 10^-7 into -log 1 x 10^-6 (if you make the 10 smaller you have to make the power bigger). So now we know that -log 3 x 10^-7 is in between -log 1 x 10^-7 and -log 1 x 10^-6. We know already know what each of the last two values are, they're 7 and 6. Therefore our answer must be between 7 and 6, and since 3 is much closer to 1 than it is to 10 (the 10 came from the -log 10 x 10^-7 that we said was -log 1 x 10^-6), we can estimate that the answer must be much closer to 7 than it will be to 6, maybe like ~6.7. And you'd most likely only have one answer choice that's very close to your estimate, in this case it was 6.5. It seems long but you'll get faster at it with practice, I'm not sure how often this would show up on the dat though.
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I think you might need it. Here's what I do. Let's say you have 2*10^-5 M H+. 1*10^-5 is a pH of 5. So 2*10^-5 is a lower pH than 5. I'd estimate it as 4.8. I would use the next power of 10 so 10^-4 and then for the decimal I do 10-2= 8 so you get 4.8. This isn't exact but it's not hard to remember and answers aren't so exact so it's basically guaranteed to get you the right answer if you pick the number closest to what you get with this method.
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Whose method did you decide on 
? Good luck! You've got this!
? Good luck! You've got this!I used your method - so great! Simple, easy, right up my alleyWhose method did you decide on
Took the test yesterday and it went great! Thanks for the help!Similar threads
