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og 12 question 11 in ps

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by aerodan1 » Thu Feb 11, 2010 10:02 pm
Hi all, I came across question 11 in the problem solving section of og 12 (page 153), and was a little confused about the explanation. the book says that the third root of 10 ^ -6 is 10 ^ -2. I suppose they divided -6 by 3 to get -2, but what is the logic behind that? thanks - dan
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by aerodan1 » Thu Feb 11, 2010 10:20 pm
ha just answered my own question, realized that the real question to ask is "what raised to the third power equals 10 ^ -6?" and of course, 10^ -2 is the answer because when numbers with the same base are multiplied by one another, their exponents are simply added. yay me
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by ajith » Thu Feb 11, 2010 11:23 pm
aerodan1 wrote:Hi all, I came across question 11 in the problem solving section of og 12 (page 153), and was a little confused about the explanation. the book says that the third root of 10 ^ -6 is 10 ^ -2. I suppose they divided -6 by 3 to get -2, but what is the logic behind that? thanks - dan
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https://www.beatthegmat.com/cubes-og-12t ... 51426.html
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by sanju09 » Fri Feb 12, 2010 12:43 am
aerodan1 wrote:Hi all, I came across question 11 in the problem solving section of og 12 (page 153), and was a little confused about the explanation. the book says that the third root of 10 ^ -6 is 10 ^ -2. I suppose they divided -6 by 3 to get -2, but what is the logic behind that? thanks - dan
Third root means cube root, and cube root of x is given by x^(1/3). When x is 10^-6, its cube root is given by (10^-6)^(1/3) and since the laws of indices say that (a^m)^n = (a)^(m n), therefore (10^-6)^(1/3) = (10)^-6(1/3) = (10)^-2.
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