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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.

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  • $\begingroup$ $x^2 = 4$ does not imply $x = 2$ - taking the positive square root of both sides is not a valid rearrangement $\endgroup$ – bounceback Jul 16 at 18:17
  • $\begingroup$ Thought the square root cancels the 2 by the X $\endgroup$ – Ashalley Samuel Jul 16 at 18:18
  • $\begingroup$ It's 16^(2x). So the 1/2 cancels the 2 $\endgroup$ – Ashalley Samuel Jul 16 at 18:19
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    $\begingroup$ It can be solved with the help of the LambertW function. $\endgroup$ – Dr. Sonnhard Graubner Jul 16 at 18:20
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    $\begingroup$ @bounceback The first operation is valid since both sides must be positive, since RHS is positive. $\endgroup$ – Allawonder Jul 16 at 18:49
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$x^{2} = 16^{2x} = (16^{x})^{2}$, 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 $-.36435$.

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

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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.

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  • $\begingroup$ "knowing that $x>0$": no. $\endgroup$ – Yves Daoust Jul 16 at 19:02
  • $\begingroup$ 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.) $\endgroup$ – usuyus22 Jul 16 at 19:31
  • $\begingroup$ Hmm thanks. Interesting $\endgroup$ – Ashalley Samuel Jul 16 at 19:41
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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).

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

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

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