Question 985654
Log[base3]x+log[basex]3=10/3
<pre>
{{{log(3,(x))+log(x,3)}}}{{{""=""}}}{{{10/3}}}

Use the base-swapping formula {{{log(A,(B))=1/log(B,(A))}}}
on the second term on the left:

{{{log(3,(x))+1/log(3,(x))}}}{{{""=""}}}{{{10/3}}}

Let u = log<sub>3</sub>(x)

{{{u+1/u}}}{{{""=""}}}{{{10/3}}}

Multiply through by LCD = 3u

      3uČ + 3 = 10u

3uČ - 10u + 3 = 0

  (u-3)(3u-1) = 0

u-3=0;  3u-1=0
  u=3;    3u=1
           u={{{1/3}}}

Substitute u = log<sub>3</sub>(x)

log<sub>3</sub>(x) = 3;   log<sub>3</sub>(x) = {{{1/3}}}

  x = 3<sup>3</sup>       x = {{{matrix(2,1,"",3^(1/3))}}}
                   
  x = 27        x = {{{root(3,3)}}}

Edwin</pre>