How to Solve for $x$ in a Particular Exponential Equation I am trying to solve for $x$ in
$x^2=(16)^{2x}.$
So I started this way:
I took square root of both sides and got
$x=16^x$
Then I took the logarithm of both sides and got
$\log x=x \log 16.$
This is where I got stuck.
 A: Contrary to some other answers,
$$x=-0.36424988978364795656$$ is the real solution. (Can be expressed in terms of the Lambert $W$ function; otherwise there is no analytical expression.)
https://www.wolframalpha.com/input/?i=x%5E2%3D256%5Ex
A: As you have stated in your question, the first step is to take both sides into a square root (knowing that $x > 0$, no roots will be lost doing this) and then to take the natural log of both sides:
$$ \sqrt {x^2} = \sqrt {16^{2x}}$$
$$ x = 16^x $$
$$ \ln x = x \ln 16 $$
$$ \frac {\ln x}{\ln 16} = x $$
Now, we know that $x$ cannot be between 0 and 1, as $-\infty < \frac {\ln x}{\ln 16} < 0$ and $0 < x < 1$. So, $x > 1$.
After this point, we will use some calculus to figure out if the lines $y=x$ and $y=\frac {\ln x}{\ln 16}$ even intersect. If we know that when $x=1$, $\frac {\ln x}{\ln 16} = 0$, we can prove that they do not intersect by finding both of their derivatives.
$$ \frac {d} {dx} [x] = 1$$
$$ \frac {d} {dx} [\frac {\ln x}{\ln 16}] = \frac {1}{x\ln16}<1$$
Because $x\ln16$ will always be greater than 1, the reciprocal of that will always be zero for $x>1$. By this, we can conclude that the second curve grows at a rate smaller than the first one for every value of x when $x>1$. Because we know the values of the first equation at $x=1$, is 1 while the other one is 0, we can say that these graphs do not intersect. So, there are no roots.
A: If one graphs both sides of this equation one gets that the real number, $x$, is about -0.3642.  Here is a Desmos graph showing this:  https://www.desmos.com/calculator/lvztuajkgk
This equation does not have an elementary solution (i.e. a solution that is able to be expressed in terms of polynomials, power functions, exponential functions, and logs). 
A: Using $x=16^x$
$x^{\frac{1}{x}}=16$
The point ($e$, ~1.4) is the "vertex" if you will, so the function will never exceed $e^{\frac{1}{e}}$ thus $x=16^x$ is undefined
A: $x^{2} = 16^{2x} = (16^{x})^{2}$, and so,
$x^{2} - (16^{x})^{2} = (x + 16^{x})(x-16^{x}) = 0.$
Now, setting $y_{1} = x + 16^{x}$ and $y_{2}=x-16^{2}$, and then graphing both equations, it will be seen that $y_{2}$ does not cross the $x$-axis (and so admits no solution to this problems), whereas $y_{1}$ crosses the $x$-axis exactly once in the interval $(-1, 0)$.
Thus, we may apply, for example, Newton's method to $y_{1} = x + 16^{x}$ to approximate the unique solution to the given exponential equation, which as somebody has already pointed out, is on the order of $-.36435$.  
