Question

What values of c and d make the equation true? 3√162x^cy^5=3x^2y(√6y^d)

Answer 1

∛(162
`x^(c) y^(5)`)=3
`x^(2) y`∛(6
`y^(d)`)
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.

Answer 2

Answer:- c=6 ; d=2

Explanation:-

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

`\\\\\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*3)y^3(6y^d)......[(a^m)^n=a^(mn)]\\\\\Rightarrow162x^cy^5=162x^6y^(3+d).........[a^(m)*\ 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`

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