Question

If 4^2x/2x+y =32 and 9^x+y/3^4y = 81 find the value of x and y

Answer 1

Given:
4^(2x/2x+y) = 32 ----(1)
9^(x+y)/3^(4y) = 81 ----(2)

Let's simplify equation (1):
4^(2x/2x+y) = 32
2^(4x/2x+y) = 2^5
4x/2x+y = 5
4x = 5(2x + y) ----(3)

Similarly, let's simplify equation (2):
9^(x+y)/3^(4y) = 81
3^(2x+2y)/3^(4y) = 3^4
2x+2y = 16
x+y = 8 ----(4)

Now we have two equations (3) and (4) with two variables x and y. Let's solve them simultaneously:

4x = 5(2x + y)
4x = 10x + 5y
6x = -5y ----(5)

x+y = 8
6x+6y = 48 ----(6)

Substituting equation (5) in (6), we get:

6(-5y) + 6y = 48
-24y = -48
y = 2

Substituting y = 2 in equation (5), we get:

6x = -5(2)
x = -5/6

Therefore, the value of x is -5/6 and the value of y is 2.

Answer 2

x= 1.25 and y = 0.75.

Explanation:

Equation 1: (4^(2x))/(2x+y) = 32

We can rewrite 32 as 2^5:

(4^(2x))/(2x+y) = 2^5

Now, we can rewrite 4^(2x) as (2^2)^(2x) = 2^(4x):

(2^(4x))/(2x+y) = 2^5

Since the bases are equal, we can equate the exponents:

4x = 5

Dividing both sides by 4:

x = 5/4 = 1.25

Equation 2: (9^(x+y))/(3^(4y)) = 81

We can rewrite 81 as 3^4:

(9^(x+y))/(3^(4y)) = 3^4

Now, we can rewrite 9^(x+y) as (3^2)^(x+y) = 3^(2x+2y):

(3^(2x+2y))/(3^(4y)) = 3^4

Again, since the bases are equal, we equate the exponents:

2x + 2y = 4

Dividing both sides by 2:

x + y = 2

We have two equations:

x = 1.25

x + y = 2

Substituting the value of x into the second equation:

1.25 + y = 2

Subtracting 1.25 from both sides:

y = 2 - 1.25 = 0.75

Therefore, the values of x and y that satisfy the given equations are x = 1.25 and y = 0.75.