As a sequel to my question How to solve $0.5^{1200}\times (2^{1204})$? :
$\frac{2^{900}*7^{898}}{14^{897}}$
Will I first solve the upper raw like did in previous question and then anwser $14^{897}$
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First of all, this has nothing to do with parsing order. You are asking for a faster way of computing it, which has nothing to do with parsing order and everything to do with factorization and various properties of multiplication/division. – Thomas Andrews Oct 11 '14 at 18:16
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Sry, by parsing I meant the calculation order of solving this one... english is not my strongest... – Kasperi Koski Oct 11 '14 at 18:18
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Okay. For what it is worth, "parsing" is how you read something, not how you do something. – Thomas Andrews Oct 11 '14 at 18:19
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$$\frac{2^{900}\cdot 7^{898}}{14^{897}} = \frac{2^{900}\cdot 7^{898}}{(2\cdot 7)^{897}} = \frac{2^{900}\cdot 7^{898}}{2^{897}\cdot 7^{897}}$$ $$= 2^{900 - 897}\cdot 7^{898-897} = 2^3\cdot 7^1 = 8\cdot 7 = 56$$
amWhy
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Check your equation again... I and the textbook got 64... maybe me and the textbook were both wrong... – Kasperi Koski Oct 11 '14 at 19:27
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My answer is correct, given the expression you posted. Perhaps you made a typo on exponents in your post? – amWhy Oct 11 '14 at 20:22
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No, there must be error in my textbook. I've alread found one with my teacher... Thanks @amWhy – Kasperi Koski Oct 12 '14 at 10:03