Radical expressions are among the most difficult topics in high school math. Let’s explore an example:<\/p>\n
√(x)7<\/sup>(5<\/sup>√(x)9<\/sup>)<\/p>\n Most people would be uncomfortable about this question. It’s really not that hard, as long as you know the exponent law<\/p>\n a<\/sup>√xb<\/sup>=x^(b\/a)<\/p>\n where a, if not written, is meant to be 2.<\/p>\n We rewrite the expression as follows:<\/p>\n √(x)7<\/sup>(5<\/sup>√(x)9<\/sup>)=x^(7\/2)x^(9\/5)<\/p>\n The law for multiplying expressions with the same base is to add the exponents. To add these fractional exponents, we need to get a common denominator as follows:<\/p>\n 7\/2 + 9\/5 = 35\/10 + 18\/10 = 53\/10<\/p>\n Therefore, we have<\/p>\n √(x)7<\/sup>(5<\/sup>√(x)9<\/sup>)=x^(7\/2)x^(9\/5)=x^(53\/10)<\/p>\n To simplify, we reverse the exponent law for mutliplication:<\/p>\n x^(53\/10)=x^(50\/10)*x^(3\/10)=x^5*x^(3\/10)<\/p>\n Reversing the radical-to-exponent law, we arrive at<\/p>\n x^5*x^(3\/10)=x^5(10<\/sup>√(x)^3)<\/p>\n The answer might seem surprising. In fact, for many, this process might need still further illumination. In future posts, I’ll revisit some of the ideas used in this one. Cheers:)<\/p>\n