Solving the system $a^x=4x+1$, $a^y=7y+6$, $x+y=5$

Question

How can I solve this exponential equation?

$$ a>1 $$ $$ a^x=4x+1 $$ $$ a^y=7y+6 $$ $$ x+y=5 $$ $a,x,y$ are real numbers

One answer is $$ a=3, x=2, y=3$$

How can I solve this?

Help me please.

Answer 1

As such, the first two equations would require the use the Lambert W function in order to tackle the combination of the exponentials with the linear terms. However, in the present case, we are able to get rid of the exponential with the help of the following trick. Indeed, one has $a^{x+y} = (4x+1)(7y+6)$ by multiplying the two equations and $a^5 = (4x+1)(7(5-x)+6)$ since $x+y = 5$.

Now, we face a mere quadratic equation, namely $28x^2 - 157x + a^5 - 41 = 0$, hence the two solutions $x_\pm = \frac{157 \pm \sqrt{29241 - 112a^5}}{56}$. Of course, those solutions are real only when $112a^5 \le 29241$. Finally, one has $y_\pm = 5 - x_\pm$. You can check that $a = 3$ leads to $x_+ = 2$ and $y_+ = 3$, as well as $x_- = \frac{101}{28}$ and $y_- = \frac{39}{28}$.