3 variables, 2 equations

to solve this you must have basic knowledge: $(log_x(y))^{-1}=log_y(x)$ $log_{x^z}(y)=\dfrac{log_x(y)}{z}$ $log_x(y^z)=z log_x(y)$ let $log_x(y)=z$ .then: $2z+5z^{-1}=2n-1\longrightarrow 2z^2-(2n-1)z+5=0$ $2.5z+1.5z^{-1}=n-3\longrightarrow 5z^2-(2n-6)z+3=0$ subtract these 2. $3z^2+5z-2=0\rightarrow z=3^{-1} or -2$ since $n=2.5z+1.5z^{-1}+3$ at $z=3^{-1}, n=\dfrac{25}{3}$ and at $z=-2,n=\dfrac{-11}{4}$ .adding these two we get $\dfrac{67}{12}$ .and $67+12=79$

Let us first perform a change-of-base to both LHS terms of the second given equation to end up with: $\frac{1}{2}log_{x} y^5 + \frac{1}{2} log_{y} x^3 = n-3$ .

Now let us substitute $u = log_{x}y, v = log_{y}x$ so that we end up with the following 2x2 system of equations:

$2u + 5v = 2n-1$

$5u + 3v = 2n-6$

which the solution is (after deploying Cramer's Rule ): $u = \frac{4n-27}{19}, v = \frac{6n+7}{19}$ . Finally, we have the system of equations:

$y = x^{(4n-27)/19}; x = y^{(6n+7)/19}$

which we can solve for $n$ according to:

$x = (x^{(4n-27)/19})^{(6n+7)/19}$ ;

or $1 = \frac{24n^2 - 134n -189}{19^2}$ ;

or $0 = 24n^2 - 134n - 550$ ;

or $n = \frac{134 \pm \sqrt{134^2 - 4(24)(-550)}}{48} = \frac{134 \pm 266}{48} = \boxed{\frac{400}{48}, -\frac{132}{48}}$

which sum to $\frac{268}{48} = \boxed{\frac{67}{12}}.$