# MCAT MATH

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#### mcatbehavior

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Does anyone have any notes on mcat math? For example, finding the antilog of 3.82without a calculator? I have tried leah4sci videos, but I am not understanding.

With questions that require such a calculation, you likely will not need to be exact. You just need to be able to estimate well enough to eliminate the other answer choices. So for this specific instance, you should know how to find an antilog in general. In other words, log(x) = 2 would mean that x = 100 because this is a base 10 log. This much should be obvious to you. Therefore, the antilog of all values between 3 and 4 lie between the antilogs of 3 and 4. This would mean log(x) = 3 and log = 4, where x = 1000 and y = 10000. So it's between 1000 and 10000.

Does anyone have any notes on mcat math? For example, finding the antilog of 3.82without a calculator? I have tried leah4sci videos, but I am not understanding.

I recommend understanding math shortcuts using logs, and memorizing the logs of 1,2,3,5,7. From there it's easy to do the math to find the others.

Log(1) = 0 of course.
Log(2) = 0.3
Log(3) = 0.477
Log(5) = 0.7
Log(7) = .85

What if we wanted to find log(6)
Well log(a*b) = log(a) + log(b) right?

So log(6) = log(2*3) = log(2) + log(3) = 0.3 + 0.477 = 0.777

Anti-log of 0.5 is 3.16
Log(1/n) = log(1) - logn = 0 - logn
So log(1/2) = -.3 (opposite of above)
Lnn = e*logn where e ~ 2.3
So ln(2)=2.3*log(2)= 2.3*.3 = 0.69

Let's try and calculate the [H+] of a solution that is pH = 4.5

log(H+) = -4.5
[H+] antilog -4.5

Hmm what could we do here? Well we know anti-log of 0.5 is 3.16. So actually

1*10^-4.5 = 3.16 * 10^-5

What if we wanted to calculate the pH of a solution that has [H+] = 7*10^-4

Long way:
-Log(7*10^-4) = -log(7) + -log(10^-4)
= -0.85 + 4 = 3.15
Shortcut:

pH = 4-log(7) = 4-0.85 = 3.15

The other quick shortcut to know is that if [H+] is under 3.16 * 10^-5 for example, then your pH is going to be between 4.5 and 5.

If [H+] is over 3.16 * 10^-5, then your pH is going to be between 4 and 4.5.

This holds true for 3.16 * 10^-whatever

I recommend understanding math shortcuts using logs, and memorizing the logs of 1,2,3,5,7. From there it's easy to do the math to find the others.

Log(1) = 0 of course.
Log(2) = 0.3
Log(3) = 0.477
Log(5) = 0.7
Log(7) = .85

What if we wanted to find log(6)
Well log(a*b) = log(a) + log(b) right?

So log(6) = log(2*3) = log(2) + log(3) = 0.3 + 0.477 = 0.777

Anti-log of 0.5 is 3.16
Log(1/n) = log(1) - logn = 0 - logn
So log(1/2) = -.3 (opposite of above)
Lnn = e*logn where e ~ 2.3
So ln(2)=2.3*log(2)= 2.3*.3 = 0.69

Let's try and calculate the [H+] of a solution that is pH = 4.5

log(H+) = -4.5
[H+] antilog -4.5

Hmm what could we do here? Well we know anti-log of 0.5 is 3.16. So actually

1*10^-4.5 = 3.16 * 10^-5

What if we wanted to calculate the pH of a solution that has [H+] = 7*10^-4

Long way:
-Log(7*10^-4) = -log(7) + -log(10^-4)
= -0.85 + 4 = 3.15
Shortcut:

pH = 4-log(7) = 4-0.85 = 3.15

The other quick shortcut to know is that if [H+] is under 3.16 * 10^-5 for example, then your pH is going to be between 4.5 and 5.

If [H+] is over 3.16 * 10^-5, then your pH is going to be between 4 and 4.5.

This holds true for 3.16 * 10^-whatever

This is a lot more complicated than needed and requires memorizing certain values.

With questions that require such a calculation, you likely will not need to be exact. You just need to be able to estimate well enough to eliminate the other answer choices. So for this specific instance, you should know how to find an antilog in general. In other words, log(x) = 2 would mean that x = 100 because this is a base 10 log. This much should be obvious to you. Therefore, the antilog of all values between 3 and 4 lie between the antilogs of 3 and 4. This would mean log(x) = 3 and log = 4, where x = 1000 and y = 10000. So it's between 1000 and 10000.

Yeah approximations are key to quickly solving problems involving calculations. EK manuals do a great job explaining this in detail.

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Does anyone know of any resources to practice mcat math?