Answer:
The equivalent will be:
[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)[/tex]
Therefore, option 'a' is true.
Step-by-step explanation:
Given the expression
[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}[/tex]
Let us solve the expression step by step to get the equivalent
[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}[/tex]
as
[tex]\sqrt[7]{x^2}=\left(x^2\right)^{\frac{1}{7}}[/tex] ∵ [tex]\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}[/tex]
[tex]\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0[/tex]
[tex]=x^{2\cdot \frac{1}{7}}[/tex]
[tex]=x^{\frac{2}{7}}[/tex]
also
[tex]\sqrt[5]{y^3}=\left(y^3\right)^{\frac{1}{5}}[/tex] ∵ [tex]\mathrm{Apply\:radical\:rule}:\quad \sqrt[n]{a}=a^{\frac{1}{n}}[/tex]
[tex]\mathrm{Apply\:exponent\:rule:\:}\left(a^b\right)^c=a^{bc},\:\quad \mathrm{\:assuming\:}a\ge 0[/tex]
[tex]=y^{3\cdot \frac{1}{5}}[/tex]
[tex]=y^{\frac{3}{5}}[/tex]
so the expression becomes
[tex]\frac{x^{\frac{2}{7}}}{y^{\frac{3}{5}}}[/tex]
[tex]\mathrm{Apply\:exponent\:rule}:\quad \:a^{-b}=\frac{1}{a^b}[/tex]
[tex]=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)[/tex] ∵ [tex]\:\frac{1}{y^{\frac{3}{5}}}=y^{-\frac{3}{5}}[/tex]
Thus, the equivalent will be:
[tex]\frac{\sqrt[7]{x^2}}{\sqrt[5]{y^3}}=\left(\:x^{\frac{2}{7}}\right)\left(y^{-\frac{3}{5}}\right)[/tex]
Therefore, option 'a' is true.
