Easy way of taking cubed roots?

[bustedgrill]

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pg 281 in Kaplin BB, KSP problems. Does anyone have a shortcut to solve these problems? I kind of suck at finding cubed roots of random numbers.

 

[joonkimdds]

can you tell us what it is?

 

[Kneecoal]

ermm... with Ksp problems you shouldn't be having to find the cubed root of anything, i think.

 

[bustedgrill]

OKay not specificly this question just KSP questions like it. For example,

what are the concntrations of each of the ions in a saturated solution of PbBr2, given the Ksp of PbBr2 = 2.1x10^-6? If 5 g of PbBr2 are bla bla bla bla....

WHat I am having trouble with is this, here is the problem set up-

2.1x10^-6=4X^3

when I solve for X I have to take the cubed root of X^3 don't I? Sorry, the easy math has escaped me.

 

[Streetwolf]

OKay not specificly this question just KSP questions like it. For example,

what are the concntrations of each of the ions in a saturated solution of PbBr2, given the Ksp of PbBr2 = 2.1x10^-6? If 5 g of PbBr2 are bla bla bla bla....

WHat I am having trouble with is this, here is the problem set up-

2.1x10^-6=4X^3

when I solve for X I have to take the cubed root of X^3 don't I? Sorry, the easy math has escaped me.

First do 2.1 x 10^-6 / 4 and get 0.525 x 10^-6. It's always best to focus on the exponent term first or else you'll have trouble later. Since it's easy to take the cube root of 10^-6, we now look at the rest of it. It's less easy to take the cube root of 0.525. I'd much rather work with something greater than 1. So how do we do that AND keep a good exponent? Since we're dealing with cube root, we need to work with 10^3 here. Any 10^(multiple of 3) will be good.

Here I'd change 0.525 into 525 by multiplying by 10^3. To counter that, we need to multiply the exponent by 10^-3. Now we have:

525 * 10^-9

This is very easy to do. The exponent becomes 10^-3. The integer is just over 8. Why? Because 8^3 = 512.

Our answer is close to 8 * 10^-3. Just a little higher.