[tex]\sqrt[3]{2}[/tex] × [tex]2^{-\frac{1}{13} }[/tex] × [tex]\sqrt[12]{32}[/tex] = [tex]2^{\frac{35}{52} }[/tex]
Applying law of indices
Applying the law of indices to the roots, we have that
[tex]\sqrt[n]{x} = x^{\frac{1}{n} }[/tex]
So,
[tex]\sqrt[3]{2}[/tex] × [tex]2^{-\frac{1}{13} }[/tex] × [tex]\sqrt[12]{32}[/tex] = [tex]2^{\frac{1}{3} }[/tex] × [tex]2^{-\frac{1}{13} }[/tex] × [tex]32^{\frac{1}{12} }[/tex]
= [tex]2^{\frac{1}{3} }[/tex] × [tex]2^{-\frac{1}{13} }[/tex] × [tex](2^{5} )^{\frac{1}{12} }[/tex]
= [tex]2^{\frac{1}{3} }[/tex] × [tex]2^{-\frac{1}{13} }[/tex] × [tex]2^{\frac{5}{12} }[/tex]
Since we have the same base in all three numbers, we have that from the law of indices, [tex]a^{x}[/tex] × [tex]a^{y}[/tex] = [tex]a^{x + y}[/tex]
= [tex]2^{\frac{1}{3} }[/tex] × [tex]2^{-\frac{1}{13} }[/tex] × [tex]2^{\frac{5}{12} }[/tex]
[tex]= 2^{\frac{1}{3} + (-\frac{1}{13}) + \frac{5}{12} } \\= 2^{\frac{1}{3} - \frac{1}{13} + \frac{5}{12} }[/tex]
Taking L..C.M of the denominators of the exponents, the L.C.M is 156.
So,
[tex]2^{\frac{1}{3} - \frac{1}{13} + \frac{5}{12} }\\= 2^{\frac{52 - 12 + 5 X 13}{156} } \\= 2^{\frac{52 - 12 + 65}{156} }\\= 2^{\frac{40 + 65}{156} }\\= 2^{\frac{105}{156} }[/tex]
Simplifying the exponent, we have
[tex]2^{\frac{105}{156} } \\= 2^{\frac{35}{52} }[/tex]
So, [tex]\sqrt[3]{2}[/tex] × [tex]2^{-\frac{1}{13} }[/tex] × [tex]\sqrt[12]{32}[/tex] = [tex]2^{\frac{35}{52} }[/tex]
Learn more about indices here:
https://brainly.com/question/10339517
