When it comes to mathematical calculations on the MCAT shortcuts are the way to go. But with most shortcut you have the option of falling back on the long yet proven pen-and-paper calculations to get your results.
But this is not the case when it comes to logarithms. Logs or negative logs, unlike multiplication, division and similar cannot be solved in a simple written out manner.
Therefore, learning a non-calculator trick for solving log questions is a must for every MCAT student. And that's why my newest video shows you how to quickly tackle simple negative log questions, and a trick for solving the more difficult questions, all without a calculator.
Logarithms and Negative Logs Without A Calculator
(click to watch on YouTube or catch the video transcript here)
MCAT Style Question Mentioned In This Video:
Chemistry Question: Find the pH of a 2.3 x 10¯⁴ molar solution of NaOH
This is Video #8 in my MCAT Math Without A Calculator tutorial series. Click HERE for the entire series
Erick B says
For these type of problems, it’s a lot quicker to use m-0.n. -log (2.3 * 10^-4) = 4 – 0.23 (simply move the decimal one to the left) = 3.77.
pH + pOH = 14, so pH = 14 – 3.77 = 10.33
Do you have a video for positive exponential logarithms? Or is this something that we should not worry about for the MCAT? Does the trick only work for negative logs of negative exponents?
Pushtee S Jhaveri says
Hi Leah! Your videos are super helpful. In the example of the -log (4.5 * 10^-3), how did you know to round the 10^-3 up to 10*10^-3? I am really confused as to how you did this!
Sana Ali says
I do not quite understand how on the last question 2.3 was close to 3 so the answer must be close to 0.5 On the cheat sheet, it says to know the log range values but I am not sure why the calculator values are different from what you show in the videos & why we should know the values under the MCAT? For ex: you said 3×10 raised to -3 = 2.53 but on the cheat sheet, it says, 3×10 raised to -3= #.523–># 0.5. Could you please elaborate on this? Thank you!
Manoj gaur says
Can u tell me how to solve log(2.3×10^2) by writing it as log(2.3+10^2)
Manoj gaur says
Can u tell me how to split 2.3×10^3 an solve by writing log(2.3+10^2)
SUPER helpful! I’m loving these tricks and have come back to them frequently throughout my studies
Sam Sayed says
HI Leah, love your videos and have been using them for my MCAT prep for about two months now. Just wanted to give you a bit of a tid bit that I picked up and see if you agree with the trick. When calculating the pH of a concentration of 3.6E-6: What i do is just take the exponent and then 1/10 of the value preceding the exponent. So for the above example I would subtract .36 from 6 to get a value of 5.64. This has been working. Is it ok?
I like your tutorial. Can you please help me in simplifying 10 to the power -4.2.
Estimate and round down to 10^4
How would you find the pH give the [H+] 9 x 10^-6 M.
I am trying to apply your concept and my answer would be ~ 6 since it’s just 9 by itself.
Can you explain how you would approach something like this?
Round 9 up to 10. Did you ask the same question on YouTube?
Nikki Bell says
Very helpful, thank you. What’s the trick if I want to find the [H+] concentration? What is 10^ -10.36?
Estimate between 10^-10 and 10^-11
Hi Leah, thanks for sharing your knowledge. But I’m having trouble with one question.. how come this rule doesn’t apply to -log(1.5×10^-4)? From what I’ve understood, it should be between 3 and 4 and since 5 is #.3, shouldn’t it be 3.3?? But the answer is 3.8. am I missing something here?:S
Thank you in advance!
MJ: 1.5 is between 1 and 2, not 5
how would you ( log 4.5 X 10^5) and (log 4.5 X 10^-5)
Thanks for sharing your knowledge!
You’re welcome. Are you pre-med?
Imani Agard says
Hey Leah, I love your videos so far, but I’m having problems understanding how you make the estimations. Take for example -log (2.3* 10^-4).
What is your reasoning behind deducing that the pH will lie between 3 and 4? Why are you rounding down to one and rounding up to ten? I think you rushed on that concept and it missed me.
Thanks for your help.
The log trick only works on numbers like 1 and 10. So if you estimate that 2 lies betwen 1 and 10 both times 10^-4 you get your range
how will you find -log(1.8 X 10^ -5 ) ?
what will you do for .8 after finding the range which is between 5 and 4..?
thanks in advance..
Don’t overthink it Jagan. On the MCAT an answer that is close enough is good enough. Estimate your equation as -log(2×10^-5) or even better, 1 x 10 ^ -5
AS a sophomore university don’t you think it’s soon for me, for the MCAT Exams.?
I don’t want to go for it twice.
Cameus it depends when you want to test. I think you’re a bit early