# 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.

• $x^2 = 4$ does not imply $x = 2$ - taking the positive square root of both sides is not a valid rearrangement – bounceback Jul 16 at 18:17
• Thought the square root cancels the 2 by the X – Ashalley Samuel Jul 16 at 18:18
• It's 16^(2x). So the 1/2 cancels the 2 – Ashalley Samuel Jul 16 at 18:19
• It can be solved with the help of the LambertW function. – Dr. Sonnhard Graubner Jul 16 at 18:20
• @bounceback The first operation is valid since both sides must be positive, since RHS is positive. – Allawonder Jul 16 at 18:49

$$x^{2} = 16^{2x} = (16^{x})^{2}$$, 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 $$-.36435$$.

• Ohk. That's very understandable. Will try this one. Thanks buddy – Ashalley Samuel Jul 17 at 5:43

$$x=-0.36424988978364795656$$ 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

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.

• "knowing that $x>0$": no. – Yves Daoust Jul 16 at 19:02
• We know that $x > 0$ because in x is negative, the left side of the equation $x = 16^x$ will be negative while the right side will always be positive (Edit: you're right.) – usuyus22 Jul 16 at 19:31
• Hmm thanks. Interesting – Ashalley Samuel Jul 16 at 19:41

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).

• That's an interesting graph. Thanks – Ashalley Samuel Jul 17 at 5:52

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

• Care to read the other answers. – Yves Daoust Jul 16 at 19:14
• Ok thanks man. Grateful – Ashalley Samuel Jul 17 at 5:51