Respuesta :

Ilyes39
∛(162[tex] x^{c} y^{5}[/tex])=3[tex] x^{2} y[/tex]∛(6[tex] y^{d} [/tex])
Then 162x^c y^5=3^3 x^{2*3}y^3(6y^d)
162x^cy^5=27*6x^6y^3y^d
162x^cy^5=162x^6y^{3+d}
So c=6, d=2.
JeanaShupp

Answer:- c=6 ; d=2


Explanation:-

Given equation:-[tex]\sqrt[3]{162x^cy^5}=3x^2y(\sqrt[3]{6y^d})[/tex]

[tex]\\\\\text{Taking cube on both sides, we get}\\\\\Rightarrow(\sqrt[3]{162x^cy^5})^3=(3x^2y(\sqrt[3]{6y^d}))^3\\\\\Rightarrow162x^cy^5=(3x^2y)^3(\sqrt[3]{6y^d})^3\\\\\Rightarrow162x^cy^5=27x^{2\times3}y^3(6y^d)......[(a^m)^n=a^{mn}]\\\\\Rightarrow162x^cy^5=162x^6y^{3+d}.........[a^{m}\times\ a^n=a^{m+n}]\\\\\text{Compare the power of corresponding variables, we get}\\\\\Rightarrow\ c=6\ ;\ 3+d=5\Rightarrow\ d=5-3=2[/tex]

Thus, the values of c=6 and d=2 make the given equation true.

Otras preguntas