SOLUTION: {{{ x^(2(logx)^2)=10x^3 }}}

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Question 560393: +x%5E%282%28logx%29%5E2%29=10x%5E3+
Answer by Edwin McCravy(20086) About Me  (Show Source):
You can put this solution on YOUR website!
                  x2[log(x)]² = 10x3

Take logs of both sides:

             log{x2[log(x)]²} = log(10x3)

Use rules of logarithms:

        2[log(x)]²·log(x) = log(10) + log(x³)

               2[log(x)]³ = 1 + 3·log(x)

2[log(x)]³ - 3·log(x) - 1 = 0

let u = log(x)

             2u³ - 3u - 1 = 0

Possible rational solutions for u are ±1, ±1%2F2

Try 1:

1|2  0 -3 -1
 |   2  2 -1
  2  2 -1 -2 

No that isn't a solution, since it did not give a remainder of 0.

Try -1:

-1|2  0 -3 -1
  |  -2  2  1
   2 -2 -1  0 

Yes -1 is a solution, so we have factored

             2u³ - 2u - 1 = 0
as

     (u + 1)(2u² - 2u - 1) = 0

   u + 1 = 0    2u² - 2u - 1 = 0
       u = -1             u = %28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
                          u = %28-%28-2%29+%2B-+sqrt%28%28-2%29%5E2-4%2A%282%29%2A%28-1%29+%29%29%2F%282%2A%282%29%29+
                          u = %282+%2B-+sqrt%284%2B8+%29%29%2F4+
                          u = %282+%2B-+sqrt%2812%29%29%2F4+
                          u = %282+%2B-+2sqrt%283%29%29%2F4+
                          u = %282%281+%2B-+sqrt%283%29%29%29%2F4+
                          u = %281+%2B-+sqrt%283%29%29%2F2+
                          
The log equation u = log(x) is equivalent to the exponential
equation x = 10u

So we have three solutions:

x = 10-1, x = +matrix%282%2C1%2C%22%22%2C10%5E%28%28%281%2Bsqrt%283%29%29%2F2%29%29%29%29, x = +matrix%282%2C1%2C%22%22%2C10%5E%28%28%281-sqrt%283%29%29%2F2%29%29%29%29

In decimal approximations they are

x = 0.1,  x = 23.22872667,  x = 0.4305014278

Edwin