Algebarically solving $-2x^2 =\ln(x)$ Is it possible to solve $-2x^2 = \ln(x)$ without using a calculator?
 A: As mentioned in the comments, there is no way in terms of "elementary functions" to express the result.  Therefore, it would be difficult to obtain even a "good" approximation without the use of a calculator/computer.
There is a "special function," however, called the Lambert W Function, "W," that is defined by the equation
$$z=W(z)e^{W(z)}$$
Therefore, we can write $-2x^2=\log x$ as $x=e^{-2x^2}$ and proceed as 
$$\begin{align}
e^{-2x^2}&=x  \\\\
\left(e^{-2x^2}\right)^2&=x^2 \\\\
4e^{-4x^2}&=4x^2\\\\
4&=4x^2e^{4x^2} \\\\
x&= \frac12\,\sqrt{W(4)}
\end{align}$$
Thus, we have
$$\bbox[5px,border:2px solid #C0A000]{x=\frac12 \sqrt{W(4)}}$$
We have as a numerical approximation for $x$
$$\bbox[5px,border:2px solid #C0A000]{x \approx 0.548217079752006}$$
A: You can solve it without a calculator, but not algebraically. The World's greatest mathematicians like Gauss, Euler, Jacobi, Riemann etc. had to do without calculators. How would they have done it without wasting time with doing lengthy calculations by hand? They could have transformed the equation into one which involves analytic functions of first order. If you put $x = \sqrt{t}$, you get:
$$\log(t) = -4 t$$
If we exponentiate both sides, we get:
$$\exp(-4t) - t = 0$$
A simple inspection leads to the conclusion that there is only one real zero at approximately $t\approx 0.3$ which we can improve on e.g. using Newton's method. But as we want to do this without a calculator and without doing lengthy calculations by hand, we won't pursue this route.
Instead, let's put $f(t) = \exp(-4t) - t$, and consider all the zeroes $\alpha_n$ of this function in the complex plane. Consider the contour integral:
$$\oint_{C(R)}\frac{1}{z^n}\frac{f'(z)}{f(z)}dz$$
where $C(R)$ is a counterclockwise contour with radius R and has as its center the origin. Then this integral tends to zero for $R\rightarrow\infty$, which means that the sum of all the residues of the integrand equals zero. From this we can conclude that:
$$\sum_{k}\frac{1}{\alpha_k^n}=-n c_n$$
where $c_n$ is the coefficient of $z^n$ in the series expansion of $\log[f(z)]$. This summation will be dominated by the real zero, which lies quite close to the origin. We can easily expand $f(t)$ in series:
$$f(t) = 1 - 5 t + 8 t^2 - \frac{32}{3}t^3 + \mathcal{O}\left(t^4\right)$$
The logarithm of this is easily calculated by hand:
$$\log[f(t)]=- 5 t - \frac{9}{2} t^2 - 9t^3+\mathcal{O}\left(t^4\right)$$
We thus have:
$$\sum_n\frac{1}{\alpha_n^2} = 9$$
and:
$$\sum_n\frac{1}{\alpha_n^3} = 27$$
If we approximate these summations by pretending that only the real term contributes to them, then we find that the zero is at approximately $1/3$. which is quite a good approximation. But we can do much better by considering the asymptotic behavior of the zeroes with large modulus and subtracting the contribution they approximately make to the summation.
So, clearly it is possible to get to quite accurate approximations without using a calculator or doing tedious computations by hand. 
