Question 560393
<pre>
                  x<sup>2[log(x)]²</sup> = 10x<sup>3</sup>

Take logs of both sides:

             log{x<sup>2[log(x)]²</sup>} = log(10x<sup>3</sup>)

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, ±{{{1/2}}}

Try 1:

1|2  0 -3 -1
 |<u>   2  2 -1</u>
  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
  |<u>  -2  2  1</u>
   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 = {{{(-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
                          u = {{{(-(-2) +- sqrt((-2)^2-4*(2)*(-1) ))/(2*(2)) }}}
                          u = {{{(2 +- sqrt(4+8 ))/4 }}}
                          u = {{{(2 +- sqrt(12))/4 }}}
                          u = {{{(2 +- 2sqrt(3))/4 }}}
                          u = {{{(2(1 +- sqrt(3)))/4 }}}
                          u = {{{(1 +- sqrt(3))/2 }}}
                          
The log equation u = log(x) is equivalent to the exponential
equation x = 10<sup>u</sup>

So we have three solutions:

x = 10<sup>-1</sup>, x = {{{ matrix(2,1,"",10^(((1+sqrt(3))/2))))}}}</sup>, x = {{{ matrix(2,1,"",10^(((1-sqrt(3))/2))))}}}</sup>

In decimal approximations they are

x = 0.1,  x = 23.22872667,  x = 0.4305014278

Edwin</pre>