$$9^m + 9^n = 52$$ $$9^m -4 = 2 \cdot 9^n$$
$$3^{m-n}=?$$
Let me show what I've tried
Simpifyling the both equalities.
$$3^{2^m} + 3^{2n} = 2 \cdot 13 \tag{1}$$ $$3^{2m} -2^2 = 2 \cdot 3^{2n} \tag{2}$$
Diving the second equality by $2$ and we have
$$\frac{3^{2m} -2^2}{2} =3^{2n} \tag{3}$$
Here is where I'm stuck.
My Kindest Regards!
Putting $$ x=9^{m}, \quad y=9^{n}, $$ we have
\begin{eqnarray*} x+y&=&52\\ x-2y&=&4 \end{eqnarray*} It follows that $$ 9^{m}=x=\frac{3x}{3}=\frac{108}{3}=36, \quad 9^{n}=y=\frac{3y}{3}=\frac{48}{3}=16, $$ i.e. $$ 3^{m}=\sqrt{9^{m}}=\sqrt{36}=6,\quad 3^{n}=\sqrt{9^{n}}=\sqrt{16}=4. $$ Hence $$ 3^{m-n}=\frac{3^{m}}{3^{n}}=\frac{6}{4}=\frac{3}{2}. $$
Remark: The numbers $m$ and $n$ are obviously not integers, in fact $$ m=\log_3(6)=\frac{\ln(6)}{\ln(3)}\approx 1.63092975\ldots ,\quad n=\log_3(4)=\frac{\ln(4)}{\ln(3)}\approx 1.261859507\ldots $$