ANSWER
2b
EXPLANATION
To simplify this expression, we have to apply some of the exponents' properties. First, the square root is a fractional exponent,
[tex]\sqrt{x}=x^{1/2}[/tex]So we can rewrite the expression as,
[tex]2\sqrt{a^2b^8}(ab^3)^{-1}=2(a^2b^8)^{1/2}(ab^3)^{-1}[/tex]Then, we can distribute the exponents into the multiplication,
[tex](xy)^z=x^zy^z[/tex]In this problem,
[tex]2(a^2b^8)^{1/2}(ab^3)^{-1}=2(a^2)^{1/2}(b^8)^{1/2}(a)^{-1}(b^3)^{-1}[/tex]Exponents of exponents are multiplied,
[tex](x^y)^z=x^{yz}[/tex]In this problem,
[tex]2(a^2)^{1/2}(b^8)^{1/2}(a)^{-1}(b^3)^{-1}=2\cdot a^{2\cdot1/2}\operatorname{\cdot}b^{8\operatorname{\cdot}1/2}\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{3\operatorname{\cdot}(-1)}[/tex]Simplify if possible,
[tex]2\cdot a^{2\cdot1/2}\operatorname{\cdot}b^{8\operatorname{\cdot}1/2}\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{3\operatorname{\cdot}(-1)}=2\cdot a^1\operatorname{\cdot}b^4\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{-3}[/tex]Now, the product of two powers with the same base is equal to the base raised to the sum of the exponents,
[tex]x^y\cdot x^z=x^{y+z}[/tex]In this problem,
[tex]2\cdot a^1\operatorname{\cdot}b^4\operatorname{\cdot}a^{-1}\operatorname{\cdot}b^{-3}=2\cdot a^{1-1}\operatorname{\cdot}b^{4-3}[/tex]Solve the subtractions,
[tex]2\cdot a^{1-1}\operatorname{\cdot}b^{4-3}=2\cdot a^0\cdot b^1=2b[/tex]Hence, the simplified expression is 2b.
