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Hello Max: Have you seen this problem of a week or so ago: 3log(x - 2) = log(2x) - 3, solve for x.

Can you take a crack at it and see how far you can go? Thanks.

Guest Jun 4, 2017
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4+0 Answers

 #1
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Here is a "ridiculous" solution from Mathematica 11, IF anybody cares to go through it. There surely MUST be a shorter solution!!.

 

Solve for x:
(3 log(x - 2))/(log(10)) = (log(2 x))/(log(10)) - 3

Subtract (log(2 x))/(log(10)) - 3 from both sides:
3 + (3 log(x - 2))/(log(10)) - (log(2 x))/(log(10)) = 0

Bring 3 + (3 log(x - 2))/(log(10)) - (log(2 x))/(log(10)) together using the common denominator log(10):
(3 log(10) + 3 log(x - 2) - log(2 x))/(log(10)) = 0

Multiply both sides by log(10):
3 log(10) + 3 log(x - 2) - log(2 x) = 0

3 log(10) + 3 log(x - 2) - log(2 x) = log(1000) + log((x - 2)^3) + log(1/(2 x)) = log((500 (x - 2)^3)/x):
log((500 (x - 2)^3)/x) = 0

Cancel logarithms by taking exp of both sides:
(500 (x - 2)^3)/x = 1

Multiply both sides by x:
500 (x - 2)^3 = x

Subtract x from both sides:
500 (x - 2)^3 - x = 0

Expand out terms of the left hand side:
500 x^3 - 3000 x^2 + 5999 x - 4000 = 0

Eliminate the quadratic term by substituting y = x - 2:
-4000 + 5999 (y + 2) - 3000 (y + 2)^2 + 500 (y + 2)^3 = 0

Expand out terms of the left hand side:
500 y^3 - y - 2 = 0

Divide both sides by 500:
y^3 - y/500 - 1/250 = 0

Change coordinates by substituting y = z + λ/z, where λ is a constant value that will be determined later:
-1/250 + 1/500 (-z - λ/z) + (z + λ/z)^3 = 0

Multiply both sides by z^3 and collect in terms of z:
z^6 + z^4 (3 λ - 1/500) - (z^3)/(250) + z^2 (3 λ^2 - (λ)/(500)) + λ^3 = 0

Substitute λ = 1/1500 and then u = z^3, yielding a quadratic equation in the variable u:
u^2 - u/250 + 1/3375000000 = 0

Find the positive solution to the quadratic equation:
u = (450 + sqrt(202485))/225000

Substitute back for u = z^3:
z^3 = (450 + sqrt(202485))/225000

Taking cube roots gives (450 + sqrt(202485))^(1/3)/(10 15^(2/3)) times the third roots of unity:
z = (450 + sqrt(202485))^(1/3)/(10 15^(2/3)) or z = -(-450 - sqrt(202485))^(1/3)/(10 15^(2/3)) or z = ((-1)^(2/3) (450 + sqrt(202485))^(1/3))/(10 15^(2/3))

Substitute each value of z into y = z + 1/(1500 z):
y = (sqrt(202485) + 450)^(1/3)/(10 15^(2/3)) + 1/(10 (15 (sqrt(202485) + 450))^(1/3)) or y = (-1)^(2/3)/(10 (15 (sqrt(202485) + 450))^(1/3)) - (-sqrt(202485) - 450)^(1/3)/(10 15^(2/3)) or y = ((-1)^(2/3) (sqrt(202485) + 450)^(1/3))/(10 15^(2/3)) - 1/10 ((-1)/(15 (sqrt(202485) + 450)))^(1/3)

Bring each solution to a common denominator and simplify:
y = ((sqrt(202485) + 450)^(2/3) + 15^(1/3))/(10 15^(2/3) (450 + sqrt(202485))^(1/3)) or y = ((-15)^(2/3) - (-15)^(1/3) (sqrt(202485) + 450)^(2/3))/(150 (450 + sqrt(202485))^(1/3)) or y = -((sqrt(202485) - 450)^(1/3) - (-1)^(2/3) (sqrt(202485) + 450)^(1/3))/(10 15^(2/3))

Substitute back for x = y + 2:
x = (15^(1/3) + (sqrt(202485) + 450)^(2/3))/(10 15^(2/3) (sqrt(202485) + 450)^(1/3)) + 2 or x = ((-15)^(2/3) - (-15)^(1/3) (sqrt(202485) + 450)^(2/3))/(150 (sqrt(202485) + 450)^(1/3)) + 2 or x = 2 - ((sqrt(202485) - 450)^(1/3) - (-1)^(2/3) (450 + sqrt(202485))^(1/3))/(10 15^(2/3))

(3 log(x - 2))/(log(10)) ≈ -2.41495 + 2.759 i
(log(2 x))/(log(10)) - 3 ≈ -2.41495 + 0.030247 i:
So this solution is incorrect

(3 log(x - 2))/(log(10)) ≈ -2.36393
(log(2 x))/(log(10)) - 3 ≈ -2.36393:
So this solution is correct

(3 log(x - 2))/(log(10)) ≈ -2.41495 - 2.759 i
(log(2 x))/(log(10)) - 3 ≈ -2.41495 - 0.030247 i:
So this solution is incorrect

The solution is:
Answer: | x = (15^(1/3) + (sqrt(202485) + 450)^(2/3))/(10 15^(2/3) (sqrt(202485) + 450)^(1/3)) + 2 = 2.1629388880535156

Guest Jun 4, 2017
 #2
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I don't think there is a shorter  " exact "  solution than that.... because in order to get x all by itself on one side of the equation, you must go through the process of solving a cubic equation.

hectictar  Jun 4, 2017
 #3
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hectictar: I just programmed the equation into my computer and it came back with the answer as:

x =2.16293888805 in less a second. It used Newton's method to home in on the answer. Perhaps, it is much easier and faster to solve such equations using this Newton's method. 

Guest Jun 4, 2017

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