Fractional ExponentsDate: 12/07/2003 at 17:06:20 From: Jessica Subject: exponent When a number is raised to a power like 4/3 or 3/5, how is it done? For example, how would I evaluate 27^(4/3)? I know that raising to the 1/2 power is the same as the square root. Date: 12/07/2003 at 18:40:48 From: Doctor Schwa Subject: Re: exponent Hi Jessica, You're right, 27^(1/2) is the square root of 27. Why? Because you add exponents when multiplying like bases: 27^(1/2) * 27^(1/2) = 27^(1/2 + 1/2) = 27^1 = 27. So 27^(1/2) is the number that, multiplied by itself, makes 27. Therefore it's the square root of 27. By the same logic, 27^(1/3) must be the third or cube root of 27: 27^(1/3) * 27^(1/3) * 27^(1/3) = 27^(1/3 + 1/3 + 1/3) = 27^1 = 27 So when you cube 27^(1/3) you get 27, meaning that 27^(1/3) is in fact the cube root of 27. Another way to look at this is to remember that when you raise a power to a power, the exponents multiply. So: [27^(1/2)]^2 = 27^(1/2 * 2) = 27^1 = 27 [27^(1/3)]^3 = 27^(1/3 * 3) = 27^1 = 27 You can see how the squaring 'undoes' the 1/2 power and the cubing undoes the 1/3 power. Now, continuing with that multiplying exponent approach, let's look at your question of how to evaluate 27^(4/3). There are actually two ways to think about it: 27^(4/3) = [27^(1/3)]^4 = 3^4 = 81 or 27^(4/3) = [27^(4)]^(1/3) = (531441)^(1/3) = 81 The easiest way to think about fractional exponents is to treat the denominator as the root to take, and the numerator as the power to raise it to. So, in the example we just did, you can take the cube root of 27 and then raise that answer to the fourth power, or you can raise 27 to the fourth power and then take the cube root of that answer. In both cases you are taking a cube root and raising to the fourth power, and you can decide which step you want to do first. In most cases taking the root first is simplest because it reduces the number that you will then raise to the power. Here are some more examples: 16^(5/2) = [16^(1/2)]^5 = 4^5 = 1024 16^(3/4) = [16^(1/4)]^3 = 2^3 = 8 Does that help clear things up? - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
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