Math calculation question

[SayAahh]

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I'm having a hard time figuring out an efficient way of calculating some things. For example,

2.1 x 10^-6 = 4(x)^3


I know to divide 2.1.... by 4 to get

5.25 x 10^-7 = x^3

Cube root of 10^-7 is (10^-7)^(1/3) and you multiply the exponents, fair enough. But now how do I figure out cube root of 5.25 quickly?

Or is there an altogether faster approach to such calculations?

 

[jntruong2003]

hey sayahhh,

so you are trying to find the cube of 5.25 x 10^-7.
what i do is i find the number that is cubed that will resemble the first number in 5.25, which in this case the number 5. for example, i know that 5^3= 125 and that answer does not start with the number 5. so through quick math i find that 8^3 is 512, which does start with the number 5. now, i know that the answer will be some form of answer w/ 8 in it. i need 8^x to equal 10^-7 at this point. b/c 8^3 is 512 that is 100 times larger or 10^2 of 5.12. so i know that 8^-3 is the answer because (10^-3)^3 is 10^-9 but because 8^3 is 512, i know i have to multiply 10^2 to 10^-9 = 10^-7. i know that it 8 will be the starting number because of 8^3 is 512. it's hard to explain, but i hope you can decipher what i'm saying.

 

[CannonD]

Sayahh,

i don't know if this is a good way, but my way is just used to estimate, but it's pretty closed.
you divide 2.1 by 4 then you would get .525x10^-6=x^3
what i did was i took Log both sides then you would get
log .525 -6log10= 3 logx and then i divided 3 to each side and get (log.525)/3 - 2 = logx (since log10=1) and you rearrange equation 10^((log.525)/3 -2) =x this you can estimate that log.525/3 is a very small value because it being divided by 3 also since (log1=0), so something small - big number, don't affect much so x=10^-2 i mean like very closed. well i don't know if this work with other problem like this.

 

[Streetwolf]

I'm having a hard time figuring out an efficient way of calculating some things. For example,

2.1 x 10^-6 = 4(x)^3


I know to divide 2.1.... by 4 to get

5.25 x 10^-7 = x^3

Cube root of 10^-7 is (10^-7)^(1/3) and you multiply the exponents, fair enough. But now how do I figure out cube root of 5.25 quickly?

Or is there an altogether faster approach to such calculations?

I'd stick with 2.1/4 x 10^-6 = x^3 instead of having 10^-5.

Then when you take the cube root you have (cube root of 2.1/4) x 10^-2.

So you want the cube root of 2.1/4 = 0.525.

Think about cubes - you have 8^3 = 512 which means 0.8^3 = 0.512. So you're REALLY close to that. Since 9^3 = 729 and 0.9^3 = 0.729, you have a long ways to go before you hit 0.9 as your answer. So I'd go with something pretty close to 0.8, maybe not even 0.81 -- go with 0.805 to be safe. The actual answer is ~0.8067 so we're good. With 0.805 x 10^-2 you get 0.00805 for an answer.

Actual answer is ~0.008067.