How do I solve $5^x = x^2$? How do I solve the equation in the title?
This was in a Calculus text book that did not show the answer to this.
 A: If $x \ge 0$ then $5^x > x^2$, and if $x \le -1$ then $5^x < x^2$, as can be proved by considering derivatives.
Now $5^{-1} = \frac{1}{5} < 1 = (-1)^2$, and $5^0 = 1 > 0 = 0^2$, so by the intermediate value theorem there is a solution to the equation $5^x=x^2$ with $x \in ({-1}, 0)$. Moreover, this solution is unique since on this interval $x \mapsto 5^x$ is increasing and $x \mapsto x^2$ is decreasing.
So with the knowledge that a solution $x$ to the equation $5^x=x^2$ exists, you need to find $x$. Unfortunately no closed form for its value exists, but you can use numerical methods to find what the value of $x$ is to within any fixed margin of error.
A: Is this problem in a section regarding Newton's Method of successive approximations?
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Let $f(x) = x^2 -5^x$. We want $x$ such that $f(x)=0$.
Since  $f(0) = -1$ and $f(-1)=0.8$, there appears to be a zero in the interval $x \in (-1,0)$. 

We can choose $x_1 = -1$ and apply Newton's method until reaching the desired accuracy. To use Newton's Method, we need $f'(x) = 2x - 5^x \ln5$.
$x_{2} = x_1 - \frac{f(x_1)}{f'(x_1)} = -1 - \frac{4/5}{-2 - 5^{-1} \ln5} \approx -0.65545$
$x_{3} = x_2 - \frac{f(x_2)}{f'(x_2)} \approx -0.61196$
$x_{4} = x_3 - \frac{f(x_3)}{f'(x_3)} \approx -0.61140$
$x_{5} = x_4 - \frac{f(x_4)}{f'(x_4)} \approx -0.61140$
$\boxed{x \approx -0.61140}$
Let's check...
$f(-0.61140) \approx 9.7 \times 10^{-7}$
If that's small enough for your liking, you've got your approximation. If not, use more decimal places in the steps above.
A: Actually, we can find a closed form solution in terms of the Lambert W Function.  We simply note that the $W$ function is defined as the solution to 
$$z=W(z)e^{W(z)} \tag 1$$
Then for $x<0$, starting with $5^x=x^2$ we take the logarithm of both sides and find
$$x\log 5=2 \log| x| \tag 2$$
Next, we divide both sides of $(2)$ by $2$ and use $\frac12 \log 5=\log 5^{1/2}$ to obtain
$$x\log 5^{1/2}=\log |x|\tag 3$$
whereupon exponentiation of both sides of $(3)$ reveals 
$$e^{-|x|\log 5^{1/2}}=|x| \tag 4$$
Finally, we bring the exponential and linear terms to one side and multiply both sides of the result by $\log 5^{1/2}$ to obtain
$$(|x|\log 5^{1/2})e^{|x|\log 5^{1/2}}=\log 5^{1/2} \tag 5$$
whereupon comparison to $(1)$ yields 
$$\bbox[5px,border:2px solid #C0A000]{x=-2\frac{W\left(\frac12\log 5\right)}{\log 5}\approx. -0.611399468715555024901
}$$
where we used the approximation for $W\left(\frac12\log 5\right)\approx. 0.49200474104107505485$.
A: Just to add to the solution posted by Clive. You can get an idea by plotting the curves.

A: $\bf{My\; Solution::}$ Let $$f(x) = 5^x-x^2\;,$$ Then $$f'(x) = 5^x\cdot \ln(5)-2x$$
and $$f''(x) = 5^x\cdot [\ln(5)]^2-2$$ and $$f'''(x)=5^x\cdot \cdot [\ln(5)]^3>0\forall x\in \mathbb{R}$$
So Using $\bf{IMVT}$ Theorem If $f'''(x)=0$ has no roots, Then $f''(x) =0 $ has at most $1$ roots
and $f'(x) =0$ has at most $2$ roots and $f(x)=0$ has at most $3$ roots.
Now for $f(x) = 5^x-x^2\;,$ at $x=-1\;,$ we get $\displaystyle f(-1) = -\frac{1}{5}-1=-\frac{6}{5}<0$ and $f(0)=1>0$
So one root of $f(x) =0$ lie between $(-1,0)$
Now when $x<-1\;,$ we get $f(x)<0$ and for $x>0\;,$ we get $f(x) = 5^x-x^2>0$
So there is exactly one root of $f(x) =0$ which lie between $(-1,0)$
