Let $p > 1$ and $q > 1$. Consider an equation in $x$ $$e ^ { 2 x } - a e ^ { x } + b = 0 \tag{1}$$ such that the equation in $t$ obtained by setting $t = e ^ { x }$ in (1) $$t ^ { 2 } - a t + b = 0 \tag{2}$$ has the solutions $\log _ { q ^ { 2 } } p$ and $\log _ { p ^ { 3 } } q$. We are to find the minimum value of $a$ and the solution of equation (1) at this minimum. (1) First of all, we see that $$b = \frac { \mathbf { A } } { \mathbf { A B } }$$ and $$a = \frac { \mathbf { C } } { \mathbf{D} } \log _ { q } p + \frac { \mathbf { E } } { \mathbf { F } } \log _ { p } q .$$ (2) As long as $p > 1$ and $q > 1$, it always follows that $\log _ { p } q > \mathbf { G }$. Hence, $a$ takes the minimum value $\frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$ when $\log _ { p } q = \frac { \sqrt { \mathbf { J } } } { \mathbf { K } }$. In this case, the solution of (1) is $$x = - \frac { \mathbf { L } } { \mathbf { M } } \log _ { e } \mathbf { N } .$$
Let $p > 1$ and $q > 1$. Consider an equation in $x$
$$e ^ { 2 x } - a e ^ { x } + b = 0 \tag{1}$$
such that the equation in $t$ obtained by setting $t = e ^ { x }$ in (1)
$$t ^ { 2 } - a t + b = 0 \tag{2}$$
has the solutions $\log _ { q ^ { 2 } } p$ and $\log _ { p ^ { 3 } } q$.\\
We are to find the minimum value of $a$ and the solution of equation (1) at this minimum.\\
(1) First of all, we see that
$$b = \frac { \mathbf { A } } { \mathbf { A B } }$$
and
$$a = \frac { \mathbf { C } } { \mathbf{D} } \log _ { q } p + \frac { \mathbf { E } } { \mathbf { F } } \log _ { p } q .$$
(2) As long as $p > 1$ and $q > 1$, it always follows that $\log _ { p } q > \mathbf { G }$. Hence, $a$ takes the minimum value $\frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$ when $\log _ { p } q = \frac { \sqrt { \mathbf { J } } } { \mathbf { K } }$. In this case, the solution of (1) is
$$x = - \frac { \mathbf { L } } { \mathbf { M } } \log _ { e } \mathbf { N } .$$