solve 8^x=4^2y+1 and 27^2x=9^y-3 giving your answer as an exact fraction

Answer 1

To solve the equations, we can rewrite the bases using the same exponent. For the first equation, we can rewrite 8 as 2^3 and 4 as 2^2.

So, the equation becomes:
(2^3)^x = (2^2)^(2y+1)
2^(3x) = 2^(2(2y+1))

Since the bases are equal, we can equate the exponents:
3x = 2(2y+1)

Now, let's solve the second equation.
We can rewrite 27 as 3^3 and 9 as 3^2.

So, the equation becomes:
(3^3)^(2x) = (3^2)^(y-3)
3^(6x) = 3^(2(y-3))

Again, equating the exponents:
6x = 2(y-3)

Simplifying both equations, we have a system of linear equations:
3x = 2(2y+1) ----(1)
6x = 2(y-3) ----(2)

From equation (1), we get:
3x = 4y + 2 ----(3)

Now, we can substitute equation (3) into equation (2):
6x = 2(3x - 9)
6x = 6x - 18
0 = -18

Since we have reached a contradiction (0 = -18), there is no solution to this system of equations.
Hence, there is no exact fraction solution to the given system.

Answer 2

8^x=4^(2y+1)
3x = 4y+2

27^(2x)=9^(y-3)
6x = 2y-6
so now solve
3x = 4y+2
3x = y-3

4y+2 = y-3
3y = -5
y = -5/3
so x = -14/9

Answer 3

To solve the given system of equations:

Equation 1: 3x = 4y + 2
Equation 2: 3x = y - 3

First, we can set both equations equal to each other since they are both equal to 3x:
4y + 2 = y - 3

Simplifying this equation:
4y - y = -3 - 2
3y = -5

Dividing both sides by 3, we get:
y = -5/3

Substituting this value of y into Equation 1, we can solve for x:
3x = 4(-5/3) + 2
3x = -20/3 + 2
3x = (-20 + 6) / 3
3x = -14/3

Dividing both sides by 3:
x = -14/9

Therefore, the solution to the system of equations is x = -14/9 and y = -5/3.