math trick help
- Thread starter liveoak
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I see 5 x 10^-7 is something less than 10^-6, and the third root of that is 10^-2, so the answer is something less than 10^-2.
For the second one, we have x^3 ~ 1 x 10^-21, so x ~ 10^-7.
I just try to find the closest exponential that's a multiple of whatever root.
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To find the cube root of something, you want the exponent to be divisible by 3. The -7 should either be changed to -6 by decreasing 5 to 0.5 or changed to -9 by increasing 5 to 500. I don't know the cube root of what 0.5 is but I can probably approximate the cube root of 500. 5x5x5 =125 and 10x10x10=1000 so 500 must be somewhere between 5 and 10. I'll go ahead and say 7. Therefore the cube root of 500 is 7. The cube root of 10^-9 is just 10^-3. The final answer would be 7x10^-3 M or very close to that.
You can use similar logic with 4x^3 = 3x10^-31. Isolating x^3 yields: 0.75 x 10^-31. I don't know the cube root of any decimal but I can approximate a whole number. I want to make 0.75 greater than 1 but at the same time, make sure the exponent is divisible by 3:
0.75x10^-31 = 75x10^-33
The cube root of 10^-33 is 10^-11. If 3x3x3=27 and 5x5x5=125. The cube root of 75 must be about 4 or very close to that. The final answer should be around 4x10^-11.
