Solve the simultaneous equations:log(x+1)+2logy=2log3

Question

logx+log y=log6

Answer 1

Given the equations:
1) log(x + 1) + 2log(y) = 2log(3)
2) log(x) + log(y) = log(6)

Rearranging equation 1) in exponential form, we get:
(x + 1)(y^2) = 3^2

Simplifying equation 2) using the properties of logarithms, we get:
log(xy) = log(6)

We can eliminate logarithms in equation 2) by converting it to exponential form:
xy = 6^1

So now we have the following simplified equations:
1) (x + 1)(y^2) = 9
2) xy = 6

We can solve these equations simultaneously by substitution or elimination method.

Using the substitution method:
From equation 2), we can express y in terms of x as y = 6/x. Plugging this into equation 1), we get:
(x + 1)((6/x)^2) = 9

Expanding and simplifying, we have:
(x + 1)(36/x^2) = 9
36(x + 1)/x^2 = 9
36(x + 1) = 9x^2
36x + 36 = 9x^2
9x^2 - 36x - 36 = 0
Dividing through by 9, we have:
x^2 - 4x - 4 = 0

Now we can solve for x using the quadratic formula:
x = (-(-4) ± sqrt((-4)^2 - 4(1)(-4))) / (2(1))
x = (4 ± sqrt(16 + 16)) / 2
x = (4 ± sqrt(32)) / 2
x = (4 ± 4sqrt(2)) / 2
x = 2 ± 2sqrt(2)

So x can be 2 + 2sqrt(2) or 2 - 2sqrt(2).

Plugging these values into equation 2), we can find the corresponding values of y:
For x = 2 + 2sqrt(2):
y = 6 / (2 + 2sqrt(2))
y = 3 / (1 + sqrt(2))

For x = 2 - 2sqrt(2):
y = 6 / (2 - 2sqrt(2))
y = 3 / (1 - sqrt(2))

Therefore, the simultaneous solution is:
(x, y) = (2 + 2sqrt(2), 3 / (1 + sqrt(2))) or (2 - 2sqrt(2), 3 / (1 - sqrt(2)))