Answer:
Explanation:
A) Given the below expression;
[tex]\sqrt[4]{x^3}[/tex]The above can be written as;
[tex]\sqrt[4]{x^3}=(x^3)^{\frac{1}{4}}[/tex]Recall the below law of exponent;
[tex](a^n)^m=a^{nm}[/tex]Applying the above law of exponent to the expression, we'll have;
[tex](x^3)^{\frac{1}{4}}=x^{\frac{3}{4}}[/tex]B) Given the below expression;
[tex]\frac{1}{x^{-1}}[/tex]Recall the below law of exponent;
[tex]\frac{1}{a}=a^{-1}[/tex]Applying the above law of exponent to the expression, we'll have;
[tex]\frac{1}{x^{-1}}=x^{-1(-1)}=x^1=x[/tex]C) Given the below expression;
[tex]\sqrt[10]{x^5\cdot x^4\cdot x^2}[/tex]Recall the below laws of exponents;
[tex]\begin{gathered} a^m\cdot a^n=a^{m+n} \\ (a^n)^m=a^{^{nm}} \end{gathered}[/tex]Applying the above law of exponent to the expression, we'll have;
[tex]\begin{gathered} \sqrt[10]{x^{5+4+2}}=\sqrt[10]{x^{11}}^{}^{} \\ =(x^{11})^{\frac{1}{10}}=x^{\frac{11}{10}}^{}^{} \end{gathered}[/tex]D) Given the below expression;
[tex]x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}[/tex]Recall the below law of exponents;
[tex]a^n\cdot a^m=a^{n+m}[/tex]Applying the above law of exponent to the expression, we'll have;
[tex]x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}=x^{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}=x^{\frac{1+1+1}{3}}=x^{\frac{3}{3}}=x^1=x[/tex]We can see from the above that the below expressions are equivalent because they both yield the same result as x;
[tex]\begin{gathered} A)\sqrt[4]{x^3} \\ D)x^{\frac{1}{3}}\cdot x^{\frac{1}{3}}\cdot x^{\frac{1}{3}} \end{gathered}[/tex]