solve 8^x=4^2y+1 and 27^2x=9^y-3 giving your answer as an exact fraction
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
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.
To solve these equations, we will use the properties of exponents and logarithms.
1) 8^x = 4^(2y+1)
First, let's rewrite 8 as 2^3 and rewrite 4 as 2^2:
(2^3)^x = (2^2)^(2y+1)
Using the property of exponents that (a^m)^n = a^(m * n), we have:
2^(3x) = 2^(2(2y+1))
Since the bases are the same, we can equate the exponents:
3x = 2(2y+1)
Now simplify the equation:
3x = 4y + 2
2) 27^(2x) = 9^(y-3)
Again, let's rewrite the bases:
(3^3)^(2x) = (3^2)^(y-3)
Using the property that (a^m)^n = a^(m * n):
3^(3(2x)) = 3^(2(y-3))
Equating the exponents:
3^(6x) = 3^(2y-6)
Now, set the exponents equal to each other:
6x = 2y - 6
Now we have a system of equations:
3x = 4y + 2 ...(equation 1)
6x = 2y - 6 ...(equation 2)
We can solve this system by substituting 6x from equation 2 into equation 1:
3(2y-6) = 4y + 2
Expand and simplify:
6y - 18 = 4y + 2
Subtract 4y from both sides:
2y - 18 = 2
Add 18 to both sides:
2y = 20
Divide both sides by 2:
y = 10
Now substitute the value of y into equation 1:
3x = 4(10) + 2
3x = 40 + 2
3x = 42
Divide both sides by 3:
x = 14/3
So the solutions to the system of equations are x = 14/3 and y = 10, expressed as exact fractions.
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.