Question

You are given the equation (2x^ny^2)^m = 4x^6y^4. Find two positive​ integers, m and​ n, that make the equation true.

Answer 1

`m = 2`

`n = 3`

Explanation:

Given

`(2x^ny^2)^m = 4x^6y^4`

Required

Solve for m and n

Start by opening the bracket using laws of indices

`2^mx^(n*m)y^(2*m) = 4x^6y^4`

Express 4 as 2²

`2^mx^(n*m)y^(2*m) = 2^2x^6y^4`

Compare both sides of the equation, we have:

`2^m = 2^2` --- (1)

`x^(n*m) = x^6` --- (1)

`y^(2*m) = y^4` ---- (2)

In (1)

`2^m = 2^2`

2 cancels out on both sides; so, we have

`m =2`

In (2)

`x^(n*m) = x^6`

x cancels out on both sides; so, we have

`n * m = 6`

Substitute 2 for m

`2 * n = 6`

Divide through 2

`n = 3`

In (3)

`y^(2*m) = y^4`

y cancels out on both sides; so, we have

`2 * m = 4`

Divide through 2

`m = 2`