Solve the pair of simultaneous equations
log(x+y)=0,
2logx=log(y-1)
Use your definitions of logs to change the equations:
first one:
log(x+y) = 0 or
10^0 = x+y
x+y = 1 ------> y = 1-x
second
2logx = log(y-1)
log x^2 = log(y-1)
x^2 = y-1 -----> y = x^2+1
so x^2+1 = 1-x
x^2 + x = 0
x(x+1) = 0
x=0 or x=-1
then y = 1 or y = 0
now logx is only defined for positve values of x, so after all this we conclude that there is no solution, since neither solution pair works in the second equation.
Log(x+y) = 10^0 x+y =1 2logx= Log(y+1) Logx^2 =log (y+1) x^2 =y+1
When X=2,y=-2
X=1,y=-1
Not satisfy with the working out
To solve the pair of simultaneous equations:
Equation 1: log(x+y) = 0
Equation 2: 2logx = log(y-1)
Step 1: Solve Equation 1
Since log(x+y) = 0, we can rewrite it as 10^0 = x + y.
Simplifying, we get:
1 = x + y ...(Equation 3)
Step 2: Solve Equation 2
We can rewrite Equation 2 as:
log(x^2) = log(y-1)
Since the logarithms have the same base, we can equate the arguments:
x^2 = y - 1 ...(Equation 4)
Step 3: Substitute Equation 3 into Equation 4
Substitute the value of y from Equation 3 into Equation 4:
x^2 = (1 - x) - 1
Simplifying the expression:
x^2 = -x
Step 4: Solve for x
Rearrange the equation to get x^2 + x = 0.
Factor the equation: x(x + 1) = 0.
Hence, we have two possible solutions:
x = 0 ...(Solution 1)
x + 1 = 0, so x = -1 ...(Solution 2)
Step 5: Find the corresponding values of y for each solution.
For Solution 1, x = 0. Substituting this in Equation 3:
1 = 0 + y
y = 1 ...(Solution 1)
For Solution 2, x = -1. Substituting this in Equation 3:
1 = -1 + y
y = 2 ...(Solution 2)
Therefore, the solutions to the simultaneous equations are:
(x, y) = (0, 1) and (-1, 2).