SOLUTION: for the problem: (3^√x)(5^√3x) i got 15√3x^8. please tell me if this is correct.??

Algebra ->  Radicals -> SOLUTION: for the problem: (3^√x)(5^√3x) i got 15√3x^8. please tell me if this is correct.??      Log On


   

Question 383750: for the problem: (3^√x)(5^√3x) i got 15√3x^8. please tell me if this is correct.??
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I'm assume the expression is:
root%283%2C+x%29%2Aroot%285%2C+3x%5E8%29
Your answer is not correct. Those are two different types of roots: a 3rd (or cube) root) and a 5th root. Different types of roots cannot be multiplied the way you did.

In order to multiply two roots together they have to be the same type of root. So we have to convert these two different roots into equivalent roots of the same type. The way to change roots into different types is to use fractional exponents. So first we will rewrite your expression with fractional exponents:
%28x%29%5E%281%2F3%29%2A%283x%5E8%29%5E%281%2F5%29
Now that the exponents are fractions we can make then the same type by making the denominators the same. The Lowest Common Denominator between 3 and 5 is 15. SO we will change both fractions so they have a denominator of 15:
%28x%29%5E%28%281%2F3%29%285%2F5%29%29%2A%283x%5E8%29%5E%28%281%2F5%29%283%2F3%29%29
%28x%29%5E%285%2F15%29%2A%283x%5E8%29%5E%283%2F15%29
The denominators are now the same. We can now switch back to radicals to finish. Using the fact that a%5E%28p%2Fq%29+=+root%28q%2C+a%5Ep%29 we get:
root%2815%2C+%28x%29%5E5%29%2Aroot%2815%2C+%283x%5E8%29%5E3%29
Simplifying inside the radicals we get:
root%2815%2C+x%5E5%29%2Aroot%2815%2C+27x%5E24%29
Finally we can multiply the roots together:
root%2815%2C+x%5E5%2A27x%5E24%29
Simplifying we get:
root%2815%2C+27x%5E29%29
Last of all we simplify the radical by factoring out perfect powers of 15:
root%2815%2C+x%5E15%2A27x%5E14%29
root%2815%2C+x%5E15%29%2Aroot%2815%2C+27x%5E14%29
x%2Aroot%2815%2C+27x%5E14%29
And we are finished.