Number of solutions to these equations using algebraic ways

Question

We have these equations over real numbers : $$ x^2 = 2^x $$ and $$ e^x = \ln x$$

First of all , we need number of solutions using algebraic ways and then find value of x individually in these equations .

Please Help!

Answer 1

For your first equation, the real solutions are $x=2$, $x=4$ and $x = -\frac{2}{\ln(2)} W(\ln(2)/2)$, where $W$ is the Lambert W function.

Your second equation has no real solutions.

EDIT: The Lambert W solution comes about in the following way.
Write the equation as $$ x^2 = e^{x \ln(2)} $$

Letting $x = - \dfrac{2 w}{\ln(2)}$, this becomes $$ \dfrac{4 w^2}{\ln(2)^2} = e^{-2w} $$ i.e. $$ w e^w = \pm \frac{\ln(2)}{2}$$

Now by definition the solutions of $w e^w = c$ are $w = W(c)$ where $W$ is a branch of the Lambert W function. The real solutions are given by the "$0$" or principal branch if $c \ge -1/e$ and the "$-1$" branch if $-1/e \le c < 0$.

Now $-1/e < -\ln(2)/2 < 0$, so there we have both branches, but this happens to be one of the places where $W$ has "closed-form" values: $W_0( -\ln(2)/2) = -\ln(2)$ and $W_{-1}( -\ln(2)/2) = -2 \ln(2)$, corresponding to the equations $$\eqalign{(-\ln(2)) e^{-\ln(2)} &= - \frac{\ln(2)}{2}\cr (-2 \ln(2)) e^{-2 \ln(2)} &= -\frac{\ln(2)}{2}\cr}$$ This is where the $x=2$ and $x=4$ solutions come from.

$+\ln(2)/2 > 0$, so here we have one real solution, $w = W_0(\ln(2)/2)$ (which we write as just $W(\ln(2)/2)$), and so $x = \frac{-2}{\ln(2)} W(\ln(2)/2)$.