Need to take the cube root

[furyhecla]

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In the Kaplan Inorganic Chemistry Study Guide... Chapter 9 (Solutions), there is an example problem (page 235).

Solving the problem (on molar solubility) requires taking the cube root of a number in scientific notation.

Specifically, 2.1 x 10^-6 = 4X^3. Need to solve for X.

So first step is to divide by 4:

5.25 x 10^-7 = x^3

Then what? How do you solve for this without a calculator?

I know that it is really (5.25 x 10^7)^1/3

But that doesn't really make the math much easier... still need to take the cube root of 5.25! And what is 10^7/3?

In the answer stem, they have an EXACT answer, not one that had rounding/approximations... x = 8.07 x 10^-3....

Any suggestions? Is there a trick to solving this? It seems odd that the MCAT would expect us to do so much detailed math by hand, instead of focusing on concepts.

Thanks

 

[ilovemcat]

In the Kaplan Inorganic Chemistry Study Guide... Chapter 9 (Solutions), there is an example problem (page 235).

Solving the problem (on molar solubility) requires taking the cube root of a number in scientific notation.

Specifically, 2.1 x 10^-6 = 4X^3. Need to solve for X.

So first step is to divide by 4:

5.25 x 10^-7 = x^3

Then what? How do you solve for this without a calculator?

I know that it is really (5.25 x 10^7)^1/3

But that doesn't really make the math much easier... still need to take the cube root of 5.25! And what is 10^7/3?

In the answer stem, they have an EXACT answer, not one that had rounding/approximations... x = 8.07 x 10^-3....

Any suggestions? Is there a trick to solving this? It seems odd that the MCAT would expect us to do so much detailed math by hand, instead of focusing on concepts.

Thanks

Well, this is what I would do:

I'd round it to: 2x10^-6
Divide by 4: 0.5x10^-6

The cube root of 10^-6 is 10^-2

The cube root of 0.5 is a little more difficult. The cube root of 1 is just 1. So if I had to take a guess, I'd imagine the cube root of 0.5 to be less than 1 but greater than 0.1 - probably somewhere in between. Therefore I'd approximate it to be: 0.5 x 10^-2 ===> 5x10^-3

Then I'd check to see if any answers have a number within the same order of magnitude. If there's only 1 choice, that's most likely the answer.

 

[furyhecla]

Thanks....that helps

Well, this is what I would do:

I'd round it to: 2x10^-6
Divide by 4: 0.5x10^-6

The cube root of 10^-6 is 10^-2

The cube root of 0.5 is a little more difficult. The cube root of 1 is just 1. So if I had to take a guess, I'd imagine the cube root of 0.5 to be less than 1 but greater than 0.1 - probably somewhere in between. Therefore I'd approximate it to be: 0.5 x 10^-2 ===> 5x10^-3

Then I'd check to see if any answers have a number within the same order of magnitude. If there's only 1 choice, that's most likely the answer.

 

[cartman1980]

5.25 x 10^-7 can be written as
525 x 10-9

cube root of 10^-9 = 10^-3

in school, we were taught to memorize the squares and cubes of basic numbers (1-10). 525 would fall between 8^3 (512), and 9^3 (729), so the answer has to be something that looks like 8.xxx x 10-3 🙂

 

[SKation]

5.25 x 10^-7 can be written as
525 x 10-9

cube root of 10^-9 = 10^-3

in school, we were taught to memorize the squares and cubes of basic numbers (1-10). 525 would fall between 8^3 (512), and 9^3 (729), so the answer has to be something that looks like 8.xxx x 10-3 🙂

Hey,

I was going over how to take the cubes of numbers and I was slightly confused i understand the fact that the cube is just the number multiplied with itself three times however I dont know how I would get the answer for the cube of 27 and the such. if there is a way you know how to do it thats fast and easy please let me know!

Thanks!

 

[DrknoSDN]

Rounding and estimation. MCAT math shouldn't be that detailed as long as you know which way to round it could get pretty close.

The cube of 27 and not the cube root... 27^3
I would just do 2.5 * 2.5 = 7.5 * 2.5 = 20 ish. along with 3 zeros.

19683 is real answer but 20,000 is probably close enough for mental math.
Considering i used those approximations before checking the answer it works relatively well.

But doing something like 7.5 * 2.5 i would figure that 7.5*2=15, and adding the last half (7.5*0.5) gives you 18.75. But i rounded to 20 because i was mentally doing 2.5^3 instead of 2.7.

For big numbers just chop them into decimals.
27 million cubed is doable in your head if you use scientific notation.
2.5E7 ^2 = 7.5E14... do that again, = 18.75E21 round that up to 20E21 and then move the decimal. 2E22 would be pretty close and doable in your head.
Real answer 1.96E22. Plus if you use scientific notation for Everything you get really good at approximating the multiplication and division of decimals.

If you were asking how to find the cube root of things it's the same as what cartman1980 showed above..
Convert your number into something where the exponent is divisible by 3 then divide by 3.
So cube root of 8*10^19
is just 80*10^18 and 18/3 is 6... cuberoot 80 is slightly higher then cube of 4^3 = 64.. So it's probably about 4.2*10^6
Exact answer 4.31*10^6, so close enough.
Hope it helps.

 

[mehc012]

I'd do neither...instead, go through the answer choices and cube them. It's much easier. That's the advantage of a multiple choice exam!
Actually, first I'd write out
2*2*2 = 8
3*3*3 = 27
4*4*4 = 64

So it's likely between 3 and 4, and it's going to have an extra multiple of 10, so...
100*100*100 + 1 = 10^7

So my answer will be 3<x<4 * 10^2
If there are multiple answers like that in the choices, then I'd cube those to see which fits.

 

[America]

I'd do neither...instead, go through the answer choices and cube them. It's much easier. That's the advantage of a multiple choice exam!
Actually, first I'd write out
2*2*2 = 8
3*3*3 = 27
4*4*4 = 64

So it's likely between 3 and 4, and it's going to have an extra multiple of 10, so...
100*100*100 + 1 = 10^7

So my answer will be 3<x<4 * 10^2
If there are multiple answers like that in the choices, then I'd cube those to see which fits.

Awesome tip! Thank you so much!