This question came out in the Examination for Japanese University Admission for International Students (EJU).
$$\qquad e^{2x}-ae^{x}+b=0 \cdots (eq.1)$$
$$Let \quad t={e}^{x}$$
$$\qquad {t}^{2}-at+b=0 \cdots (eq.2)$$
The roots of this quadratic equation (eq.2) are
$\qquad \log_{{q}^{2}} p,\;$ and $\;\log_{{p}^{3}} q$, where $p>1$ ,$\;$ $q>1$
Question: Solve for $x$ when the value of $a$ is smallest.
Hints: These are the hints provided. How do I use these hints to solve this question? (I only manage to find out $a$ and $b$ by using Sum of Root and Product of Root)
- By Product of Roots $\;\to\;$ $b=\log_{{q}^{2}} p \times \log_{{p}^{3}} q$
$$b=\frac{1}{6}$$
- By Sum of Roots $\;\to\;$ $a=\log_{{q}^{2}} p + \log_{{p}^{3}} q$
$$a = \frac{1}{2}{log_q p} + \frac{1}{3}{log_p q}$$
- Given that $$log_p q > A$$
$\quad$When $$\quad log_p q = \frac{\sqrt{B}}{C} \quad$$ $a$ has the smallest value of,
$$\qquad\qquad a= \frac{\sqrt{D}}{E} \quad$$ where $A,B,C,D,E$ are integers of one digit
- The root of (eq.1), $x$ is in the form of
$$\qquad x = -\frac{F}{G}{\log_e H}$$ where $F, G, H$ are integers of one digit
Hint :
Since the roots of thus equation are real, you have the condition that discriminant $\ge 0$.
Thus, the minimum value of $a$ is
$$a^2-4b \ge 0 \implies a_{min} ^2=4b =\frac 23$$
Hence $$a_{\text{min}}=-\sqrt{\frac 23}$$
Also use that $$\log_q p=\frac{1}{\log_p q}$$