If $\;\log_m mn + \log_n mn = 49$, find $(\log_m n)^{3/4} + (\log_n m)^{3/4}.$ Given that $\log_m mn + \log_n mn = 49,\;$ find the value of $$(\log_m n)^{3/4} + (\log_n m)^{3/4}.$$
I have rewritten this expression but have hit a dead end. 
 A: Let $\sqrt[4]{\log_nm}=x$.  Hence, $x>0$ and  $$ x^4+\frac{1}{x^4}=47 \iff
\left(x^2+\frac{1}{x^2}\right)^2=49$$ 
$$\iff x^2+\frac{1}{x^2}=7$$ 
$$\iff\left(x+\frac{1}{x}\right)^2=9$$ 
$$\iff x+\frac{1}{x}=3.$$
Thus, $$\left(\log_m n\right)^{3/4} + \left(\log_n m\right)^{3/4}=x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3\left(x+\frac{1}{x}\right)=27-9=18.$$
A: Edit-new answer as original question has been modified
Using simple laws of logarithm, your equation becomes:
$$\log_m  m + \log_m n + \log_n m + \log_n n = 49$$
$$ \log_m n + \log_n m = 47$$
Using $\log_b a = \frac{1}{\log_a b}$,
You can rewrite the above equation as:
$$ \log_m n + \frac{1}{\log_m n} = 47$$
The above is a quadratic which should help you solve $(\log_m n)^\frac{3}{4} + (\log_n m)^\frac{3}{4}$
Not trivial to solve using this method-see Michael Rozenberg's answer. 
A: From the condition we get $$\log_m m+\log_m n+\log_n m+\log_n n=49$$
thus we have $\log_m n+\log_n m=47.\,$ Setting $t=\log_m n$, we get
$t+\frac{1}{t}=47.$
Solving this we get $$t_1={\frac {47}{2}}+21/2\,\sqrt {5}\iff t_2={\frac {47}{2}}-21/2\,\sqrt {5}$$
Can you finish?

I'll finish for you: Plugging this in your term we get $$\left( {\frac {47}{2}}+21/2\,\sqrt {5} \right) ^{3/4}+ \left( {\frac 
{47}{2}}-21/2\,\sqrt {5} \right) ^{3/4}=18
$$
