Solve #4x^(1//2)#?

1 Answer

#4x^"1/2" = 4 sqrt(x)#

Explanation:

Assuming your question is #4x^"1/2"#, we can use the laws of exponents to solve.

Rule: #x^"m/n"=root(n)(x^m)#
Problem: #4x^"1/2"#
#= 4*x^"1/2"#
#= 4*root2(x^1)#
# = 4sqrt(x)#


Note that if the question you meant to ask is #(4x)^"1/2"# then you can use the laws of exponents to solve as follows.

Rule: #x^"m/n"=root(n)(x^m)#
Problem: #(4x)^"1/2"#
#=root2((4x)^1)#
#=sqrt(4x)#

Rule: #sqrt(ab)=sqrta*sqrtb#
#sqrt(4x)=sqrt(4) * sqrt(x)#
#=2sqrtx#

More:
https://www.mathsisfun.com/algebra/exponent-laws.html