Math tip:

[the_fella]

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I just encountered a problem where I had to take the square root of 600,000. I was trying to figure out how many zeros go after the answer (which I had already approximated), when it dawned on me, I can convert it to scientific notation (with an even exponent) and do it that way.

sqrt(60*10^4) = ~ 7.8*10^2, so ~ 780

Btw, I approximated the sqrt of 60 by knowing that it's between 49 (7*7) and 64 (8*8), but closer to 8. So approximately 7.8

I hope someone finds this helpful.

 

[NextStepTutor_3]

I'm so glad you posted this! It's a great trick. The key, like you mentioned, is the even exponent. I've always used "Find the square root of 0.0064" as a simple example. We can either struggle to figure out whether it's .08. .008, etc., or we can convert to scientific notation. Finding the square root of 64 x 10^-4 is a piece of cake - take the square root of the coefficient, giving us 8, and cut the exponent in half, leaving us with a final answer of 8 x 10^-2 or 0.08.

This can also work if the number is already in scientific notation, but with an odd exponent! For example, finding the square root of 8.1 x 10^7 looks heinous at first glance. But it's much easier if we make it 81 x 10^6 ! It just goes to show how scientific notation is meant to make our lives easier, no matter how often it may seem to be the opposite 🙂

 

[the_fella]

Fwiw, the laws of logs are something else to remember, especially for pH, pKa, and whatnot, especially if you remember the logs of 1, 3, 5, 7, and 10. That way, if you're expected to approximate the log of 4, you can use the law of logs to do so: 4=2*2. log 2*2 = log 2 + log 2 = log 4 = 0.6


log A + log B = log (A*B)

log A - log B = log (A/B)

log (A)B = B(log A)

Also, the trick I mentioned in my previous post for finding the square root also works for the cube root. For instance, if you're trying to find the cube root of 25.
2^3 = 8
3^3 = 27

So you know your answer must be between 2 and 3, but since 25 is closer to 27 than to 8, we know the answer is much closer to 3. So about 2.8 or so would be a reasonable approximation. It's actually 2.92, so pretty close!