Calculating 10^(.1)

[shaq786]

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how would you calculate 10^(.1) by hand on the mcat? or 10^(.2)

 

[Daichi Katase]

where did you encounter a need to do that???

 

[Turkeyman]

wow lol

wouldnt that be 10^(1/10) or the tenth root of 10?

Kinda like 10^(1/2) or 10^(1/3) is the square root/cube root, respectively

That's crazy though, what number when multiplied by itself 10 time gives you 10 ;0

and why do you have to do this?

 

[stoleyerscrubz]

That's what it would be. But I don't think I've ever seen anything less than a 3rd root in my life.

wouldnt that be 10^(1/10) or the tenth root of 10?

Kinda like 10^(1/2) or 10^(1/3) is the square root/cube root, respectively

QUOTE]

 

[QofQuimica]

how would you calculate 10^(.1) by hand on the mcat? or 10^(.2)

You should never run into these kinds of calculations on the MCAT; they know that you don't have a calculator, and no one expects you to know the tenth or fifth roots of numbers! If by some bizarre chance you did get a problem like that, you should try to estimate the answer by bracketing it. For example, I know that 1^10 = 1, and 2^10 = 1024. So the tenth root of 10 will be a number between 1 and 2, closer to 1 than to 2. (maybe 1.1 ish?) 10^.2 is the same thing as the fifth root of 10, which would also be a number between 1 and 2 (1^5 = 1 and 2 ^5 = 32, so I'd guess something like 1.5 ish). In general, never waste time on the MCAT doing exact calculations. Round and take short cuts with the math as much as you possibly can.

 

[Reimat]

My solution would be make a good guess- I can't imagine this actually being on a test though.

Logically, it's pretty quick to come up with a reasonable answer- you know it has to be >1, <2. Then, if you just start thinking about numbers^10 between 1 and 2, you can get a reasonable approximation.

 

[Reimat]

😀 haha, I'm guessing QofQuimica and I were thinking along the same lines while responding to the post at the same time