Question 1206837
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Find x, {{{3^x}}} = {{{x^9}}}.
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<pre>
Your starting equation is 

    {{{3^x}}} = {{{x^9}}}.    (1)


Take natural logarithm of both sides

    x*ln(3) = 9*ln(x).


Divide both sides by ln(x)*ln(3).  You will get

    {{{x/ln(x)}}} = {{{9/ln(3)}}}.


Transform right side equivalently multiplying by 1 = {{{3/3}}}

    {{{x/ln(x)}}} = {{{(3/3)*(9/ln(3))}}} = {{{(3*9)/(3*ln(3))}}} = {{{27/ln(3^3)}}} = {{{27/ln(27)}}}.


Thus we have this equation 

    {{{x/ln(x)}}} = {{{27/ln(27)}}}.    (2)


It is well known fact that if x > e, then {{{x/ln(x)}}} is monotonically increasing function.
Here e is the base of natural logarithms, e = 2.71818...
THEREFORE, from equation (2), we conclude that x = 27 is the unique solution in domain x > e.


<U>ANSWER</U>.  In domain x > e, the unique solution to equation (1) is  x= 27.


<U>CHECK</U>.  At x= 27, left side of equation (1) is {{{3^27}}};

                  right side of equation (1) is {{{27^9}}} = {{{(3^3)^9}}} = {{{3^27}}}.

        Thus, at x= 27, both sides of equation (1) are equal.
</pre>

Solved.


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Post-solution notes


<pre>
    1. The original equation has the second solution in the domain  1 < x < e, where the function {{{x/ln(x)}}} 
       is also monotonic; but this solution is not an integer number.
       This second solution can be found numerically as an approximate value.


    2. This trick with using monotonicity of the function {{{x/ln(x)}}} is a powerful tool for solving
       many/some exponential-polynomial equations, similar to the given in this post.
</pre>