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

Answer:

$m = 2$

$n = 3$

Step-by-step 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

$m = 2$