Solve for x and y if log_2(x+y)=log_3(3x+4y)=3
To solve this problem, we will use the properties of logarithms.
First, using the property log_b(x) = log_c(x) / log_c(b), we can rewrite the equation log_2(x+y) = 3 as (x+y) = 2^3 = 8.
Next, using the property log_b(x) = log_c(x) / log_c(b), we can rewrite the equation log_3(3x+4y)=3 as (3x+4y) = 3^3 = 27.
So, we have the following system of equations:
x + y = 8 ...(1)
3x + 4y = 27 ...(2)
To solve this system, we can use the method of substitution.
Simplifying equation (1) to solve for x, we have:
x = 8 - y ...(3)
Substituting equation (3) into equation (2), we have:
3(8 - y) + 4y = 27
24 - 3y + 4y = 27
-y = 27 - 24
-y = 3
y = -3
Now substituting the value of y into equation (1), we have:
x + (-3) = 8
x - 3 = 8
x = 8 + 3
x = 11
Therefore, the solution to the system of equations is x = 11 and y = -3.
To solve for x and y, we will use the given equations:
log_2(x+y) = log_3(3x+4y) = 3
Step 1: Rewrite the logarithmic expressions as exponential equations.
Using the property of logarithms: log_a(b) = c is equivalent to a^c = b, we rewrite the given equations as:
2^3 = x + y ----(1)
3^3 = 3x + 4y ----(2)
Step 2: Simplify the exponential equations.
2^3 = 8 and 3^3 = 27, so we have:
8 = x + y ----(3)
27 = 3x + 4y ----(4)
Step 3: Solve the system of equations (3) and (4).
To solve the system, we will use the method of substitution or elimination.
Using the substitution method:
From equation (3), we can solve for y in terms of x:
y = 8 - x
Substituting y = 8 - x into equation (4):
27 = 3x + 4(8 - x)
27 = 3x + 32 - 4x
27 - 32 = -x
-5 = -x
x = 5
Substituting x = 5 back into equation (3):
8 = 5 + y
y = 8 - 5
y = 3
Therefore, the solutions for x and y are x = 5 and y = 3, respectively.