Question 1156161
<pre>
{{{x^3 + 3^x = 17}}}

Try x=1

{{{1^3 + 3^1 = 17}}}
{{{1+3 = 17}}}
{{{4=17}}}

Nope.  1 is too small.

Try x=3

{{{3^3 + 3^3 = 17}}}
{{{27+27=17}}}
{{{54=17}}}

Nope. 3 is too big.

Try x=2

{{{3^2 + 2^3 = 17}}}
{{{9+8=17}}}
{{{17=17}}}

Hurray! 2 is the solution.

When an equation has a variable both as part of an exponent
and not part of an exponent elsewhere, there is no way to solve 
the equation except by trial and error, and some systematic way 
of deciding whether you need to try a bigger or a smaller number, 
based on previous trials.

Graphing calculators can handle such problems quite well.

Edwin</pre>