I understand powers but I just don't see how you can times x by itself any times that are not whole numbers. How do you times x by itself one and a half times?

Is that just one of those things that is true or is there an easy to understand reason? Sorry, I'm just curious.

Fractional powers are equivalent to roots. x^1/2 = 2nd root of x^1 x^3/2 = 2nd root of x^3 x^5/3 = 3rd root of x^5 - Warren

The short incomprihensible answer: By definition x^(1/2)= sqrt(x) an more generally x^(1/n) is the n-root of x. To understand better why (most definitions have some kind of sense): Powers of non-natural numbers are a generalization of the powers x^n = x*x*x...x (n times). For powers we have the easily verifiable rule (for n and m natural): x^n * x^m = x^(n+m) If we divide by x^m (that is multiply by 1/x^m) on both sides we obtain only x^n. That is we have to substract m in the exponent of x^(m+n) which is multiplying by x^(-m). In that way we se that our exponent-summation-rule remains valid if we allow for negative exponents, defining x^(-m) as 1/x^m. When we only allowed positive exponents we could devide x^m (i.e x^4 for m=4) by x by decreasing the exponent (obtaining x^3, x^2, x^1) each time, but after x^1 = x we didn't now what to do. With negative exponents (and zero), we can continue applying the same rule, x^0 = x^1 / x = x^(1-1) = x/x = 1, x^(-1) = 1/x, etc. So why did I spend so much time explaining negative exponents when our problem was non-integer-exponents? Because we do the same procedure over again, with another rule which is true for whole-numbered (and especially natural numbered) exponents: (x^n)^m = x^(n*m) (verify this for some low integers n and m) For clarity I will switch to n and m being numbers, i.e n=3 and m=2, that is: (x^3)^2 = x^(3*2) Supose we now want to get rid of the 2 in the exponent on the right side, like we did for the m in the first equation. We then devide by 2 in the exponent. This is the same as multiplying with 1/2. But we also now that multiplying with a (integer at least) number in the exponent is the same thing as elevating the whole expression in that number (mathematically: x^(n*m) = (x^n)^m ). Let us now use our 1/2 instead of integer and see what happens. We elevate both sides of the equation in the power 1/2 and use that this is the same as multiplying the exponent by 1/2: ((x^3)^2)^(1/2) = (x^(3*2))^(1/2) = x^(3*2*(1/2)) = x^3 We se that elevating a number en the power 1/2 in effect, applying our rules, is the same as taking the square root of the number. In the same way we can find that multiplying the exponent with 1/n (n being an integer) is the same at taking the n'th root. An example: cuberoot(x^12) = (x^12)^(1/3) = x^(12*(1/3)) = x^4 As a summary, generalizing our two rules for powers of x, we have that x^(n/m) = m-root(x^n) = (m-root(x))^n. In your example x^1.5 = x^(3/2) = sqrt(x^3) (often written as x*sqrt(x))

Welcome to sciforums, myhr -- good to have you here. Please Register or Log in to view the hidden image! - Warren

Wesmorris: "goddamn math majors Please Register or Log in to view the hidden image!" Yea, "x^(1/2)= sqrt(x) BECAUSE I SAID SO" is much better.

you see, the smiley was to indicate i was joking with him and impressed at the long winded explanation (and likely accurate if Warren didn't rip it up, I personally didn't read it that closely) that i couldn't possibly have given. i thought it self explanitory. now, i've spelled it oot. thanks for forcing my hand you bastard. Please Register or Log in to view the hidden image!* * The preceding smiley face indicates that I'm now joking with you TOO with the whole bastardness of the situation.

There may be more intuitive ways of explaining it, but what I wrote was what came to my mind yesterday night. I remember "discovering" that sqrt(x) = x^(1/2) while trying to write a computer program for the first time. It was supposed to read in some dates for a circle or a square and return some other properties. It was quite esay to calculate the area from the radii or the sides, but as I didn't find a sqrt()-function I wasn't able to do the reverse. While trying to think backwards I suddenly got this crazy thought that elevating to 0.5 must be the opposite of elevating to the second (since elevating to the fist does nothing). I had never seen non-integer powers before, and was quite amazed when the computer didn't get hickup as I ran a program which elevated a number to 0.5. I was even more astonished when I saw that the numbers mached the ones from my calculator when I took the square root. Btw, I'm not a math major Please Register or Log in to view the hidden image!