Find x and y if 3^2x-y=1 and 16^x/4=8^3x-y?
you're pretty sloppy with the parentheses, but if you mean
3^(2x-y) = 1
16^(x/4) = 8^(3x-y)
then that is the same as
2^x = 2^(9x-3y)
x = 9x-3y
Now, since 3^0 = 1, that means
2x-y = 0
Now just solve
8x - 3y = 0
2x - y = 0
Looks like (0,0) is the only solution.
Maybe you meant to group the variables in some other way?
I dont know
To find the values of x and y, we can solve the given equations step by step.
Equation 1: 3^(2x-y) = 1
Since any number raised to the power of 0 is equal to 1, we can rewrite the equation as:
3^(2x-y) = 3^0
Therefore, we have:
2x - y = 0 ---(Equation 1)
Equation 2: 16^(x/4) = 8^(3x-y)
Since 16 is equal to 2^4 and 8 is equal to 2^3, we can rewrite the equation as:
(2^4)^(x/4) = (2^3)^(3x-y)
Using the property of exponents, we can simplify further:
2^4*(x/4) = 2^3*(3x-y)
Simplifying both sides gives us:
2^(4x/4) = 2^(3*(3x-y))
Cancelling out the common base, we get:
2^x = 2^(9x-3y)
Since the bases on both sides are equal, we can equate the exponents:
x = 9x - 3y ---(Equation 2)
Now, we have a system of two linear equations:
2x - y = 0 ---(Equation 1)
x = 9x - 3y ---(Equation 2)
We can solve this system of equations using the method of substitution or elimination.
Using substitution, we can solve Equation 1 for y:
y = 2x
Substituting y in Equation 2, we get:
x = 9x - 3(2x)
x = 9x - 6x
x = 3x
Subtracting 3x from both sides, we have:
0 = 2x
Dividing both sides by 2, we get:
x = 0
Substituting the value of x into Equation 1, we can solve for y:
2(0) - y = 0
0 - y = 0
y = 0
Therefore, the solution to the given system of equations is x = 0 and y = 0.
To find the values of x and y in the given equations, we'll solve them step by step:
Equation 1: 3^(2x-y) = 1
To simplify this equation, we can rewrite 1 as 3^0, since any number raised to the power of 0 is always 1. So, we have:
3^(2x-y) = 3^0
Since the bases are the same, the exponents must be equal:
2x - y = 0
Equation 2: (16^x) / 4 = 8^(3x-y)
To simplify the equation, we can rewrite 16 as 4^2 and 8 as 2^3:
(4^2x) / 4 = (2^3)^(3x-y)
Now, we can simplify the right side using the exponent rule, which states that (a^b)^c = a^(b * c):
(4^2x) / 4 = 2^(3 * (3x-y))
Expanding further, we have:
(4^2x) / 4 = 2^(9x - 3y)
Next, we can simplify the left side by canceling out the common factor of 4:
4^(2x - 1) = 2^(9x - 3y)
Converting both sides to the same base, we can write 4 as 2^2:
(2^2)^(2x - 1) = 2^(9x - 3y)
Now we can apply the exponent rule again, (a^b)^c = a^(b * c):
2^(4x - 2) = 2^(9x - 3y)
Since the bases are equal, the exponents must be equal:
4x - 2 = 9x - 3y
Now, we have a system of two equations:
Equation 1: 2x - y = 0
Equation 2: 4x - 2 = 9x - 3y
To solve this system, we can use substitution or elimination method. Let's use substitution:
From Equation 1, we can solve for y:
y = 2x
Substituting this value of y in Equation 2:
4x - 2 = 9x - 3(2x)
Simplifying:
4x - 2 = 9x - 6x
4x - 2 = 3x
Bringing all the x terms to one side:
4x - 3x = 2
x = 2
Now, substitute the value of x back into Equation 1 to find y:
2(2) - y = 0
4 - y = 0
Solving for y:
y = 4
Therefore, the solution for x is 2, and the solution for y is 4.