Solve simultaneously log (x-1)+2logy=2log3 and logx+logy=log6
From the first, using log rules:
log (x-1)+2logy=2log3
log (x-1)+log(y^2)=log(3^2)
log( (x-1)(y^2) ) = log 9
(x-1)y^2 = 9
from the 2nd:
logx+logy=log6
log(xy) = log6
xy = 6 or x = 6/y
now use substitution,
(6/y - 1)(y^2) = 9
6y - y^2 - 9 = 0
y^2 - 6y + 9 = 0
(y - 3)^2 = 0
y - 3 = 0
y = 3
then x = 6/3 = 2
Check with your calculator.
To solve the simultaneous equations:
log(x-1) + 2log(y) = 2log(3) ----(1)
log(x) + log(y) = log(6) ----(2)
Step 1: Use the properties of logarithms to simplify the equations. Recall that log(a) + log(b) = log(a*b), and log(a^b) = b * log(a).
Rewriting equation (1):
log(x-1) + log(y^2) = log(3^2)
log(x-1) + log(y^2) = log(9)
Combining the logarithmic terms on the left side:
log((x-1) * y^2) = log(9)
Therefore, we have:
(x-1) * y^2 = 9 ----(3)
Rewriting equation (2) using the properties of logarithms:
log(xy) = log(6)
Therefore, we have:
xy = 6 ----(4)
Step 2: Solve equations (3) and (4) simultaneously.
From equation (3):
(x-1) * y^2 = 9
Rearranging equation (4):
xy = 6
We can rewrite equation (4) as:
x = 6/y
Substituting this value of x into equation (3):
(6/y - 1) * y^2 = 9
(6 - y) * y^2 = 9y
Expanding:
6y^2 - y^3 = 9y
Rearranging:
y^3 - 6y^2 + 9y = 0
Factoring out y:
y(y^2 - 6y + 9) = 0
Simplifying:
y(y - 3)(y - 3) = 0
Thus, we obtain:
y = 0, y = 3
Step 3: Substitute the values of y back into equation (4) to find the corresponding values of x.
For y = 0:
x = 6/0 (undefined)
For y = 3:
x = 6/3 = 2
Therefore, the solution to the simultaneous equations is:
x = 2, y = 3.
To solve the system of equations simultaneously, we can use properties of logarithms to simplify the equations and then solve for the variables separately.
Let's start with the first equation:
log(x-1) + 2log(y) = 2log(3)
Using the properties of logarithms, we can rewrite this equation as:
log(x-1) + log(y^2) = log(3^2)
Applying the property of adding logarithms, we get:
log((x-1)(y^2)) = log(9)
Since the logarithm function is one-to-one, we can drop the logarithm on both sides:
(x-1)(y^2) = 9
Now let's move on to the second equation:
log(x) + log(y) = log(6)
Using the property of adding logarithms, we get:
log(xy) = log(6)
Again, dropping the logarithm on both sides:
xy = 6
Now we have two equations:
(x-1)(y^2) = 9 ---(1)
xy = 6 ---(2)
We can solve this system of equations by substitution or elimination method. Let's use the substitution method:
From equation (2), we can isolate x:
x = 6/y
Substituting this into equation (1), we get:
(6/y - 1)(y^2) = 9
Expanding and simplifying the equation:
(6 - y)(y^2) = 9y^2
6y^2 - y^3 = 9y^2
Moving all terms to one side:
0 = y^3 + 3y^2 - 6y^2
Combining like terms:
0 = y^3 - 3y^2
Factor out y^2:
0 = y^2(y - 3)
This equation gives us two possible solutions:
1) y^2 = 0
y = 0
2) y - 3 = 0
y = 3
Now let's solve for x using these values of y:
For y = 0:
From equation (2), we have:
x * 0 = 6
0 = 6
This means y = 0 does not satisfy the equation.
For y = 3:
From equation (2), we have:
x * 3 = 6
x = 2
Therefore, the solution to the system of equations is x = 2 and y = 3.
To check if this solution is correct, substitute x = 2 and y = 3 into both original equations and see if they hold true.