Fractions and exponents together can be confusing. Fractions can have exponents, and exponents can be fractions. Fractions can even have fractional exponents. Any of these can be intimidating the first time you see them, but all of them can be computed in a straightforward manner if you follow a few simple patterns.

Evaluate a fraction raised to a positive exponent as you would any exponent  as repeated multiplication. For example, (3/4)^3 = (3/4) X (3/4) X (3/4) = 9/64. For large exponents, this can get complicated, but there is an exponent rule for fractions that can simplify the process. (a/b)^n = (a^n/b^n). This means that (3/4)^3 = (3^3)/(4^3) =9/64. The standard rules for 0 and 1 exponents apply, so (3/4)^0 = 1 and (3/4)^1 = 3/4. When the exponent is negative, the fraction can be inverted to get rid of the negative sign in the exponent. (a/b)^n = (b/a)^n.

Find the value of a number raised to a fractional exponent by decomposing the fraction into two factors  a number and a fraction with one in the numerator. For example, (27)^(2/3) = (27)^(2 X (1/3)) = ((27)^(1/3))^2 = 3^2 = 9. In other words, when an exponent is 1/n it means the nth root. This is clearer if you look at why, for example, K^(1/2) is the same as the square root of K. Note that (K^(1/2)) X (K^(1/2)) = K^((1/2) + (1/2)) = K^1 = K. If K^(1/2) behaves like the square root of K, it is equal to the square root of K. The same pattern can be seen for any fractional exponent with one in the numerator.

Calculate a fractional expression with exponents scattered about the fraction using the rule (a^n/a^m) = a^(n  m). For example, (5Z^12)/(10 Z^7) = (Z^5)/2 and (5Z^2)/(10 Z^7) = 1/(2Z^5). Notice that changing the sign of the exponent moves the expression up or down across the fraction bar, so (3X^2)/5 = 3/(5X^2) and 3/(5X^2) = (3X^2)/5.
Tips & Warnings
 For all these rules, when the exponent is one it can be simply erased and when the exponent is zero, the entire expression evaluates to one  no matter how complex the expression is. (X  no matter how complex)^1 = X and (Y  no matter how complex)^0 = 1
 All of these rules  like (a^n) X (a^m) = a^(n + m) and (a^n) / (a^m) = a^(n  m)  only work because the bases are the same. So (a^n) X (b^m) = a^(n + m) or (a^n) / (b^m) = b^(n  m) only if a = b.
References
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