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We have the equation in $x$: $$x^2 = 2^x$$

We know that, by logic, if we equate the bases and the powers separately, we get $x=2$ in both the cases and thus we conclude that $2$ is the root of the equation.

But what if we don't apply that logic. Is their a mathematical way we can get the same result? Using mathematical steps instead of intuition and logic?

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    We could consider this equation as a function. Find first derivate and analyse this function. – 2017-01-08
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    What about $x=4$? – 2017-01-08
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    you could use the Lambert-W function – 2017-01-08
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    @J.M.isn'tamathematician yeah! Just realised that's also a solution. So, even if we know $x=4$, how can we show it? Thanks! – 2017-01-08
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    @Dr.SonnhardGraubner could you explain the function? Thanks – 2017-01-08
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    see here https://en.wikipedia.org/wiki/Lambert_W_function – 2017-01-08
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    There are three solutions. In addition to $2$ and $4$, there is a negative solution. – 2017-01-08
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    I know I've answered this question before on this site. Just searching "x^2=2^x" you get many hits. – 2017-01-08
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    What you are calling "logic" is not logic: in fact, it is quite incorrect! – 2017-01-08
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    It should be a scandal that so many people thing there's such a thing as "mathematically rather than logically". That is OBVIOUSLY the inevitable consequence of rules requiring mathematics to be taught to vast numbers of people who don't want to learn mathematics, conjoined with the idea that the way to teach mathematics is to reqiore each student to spend a thousand years mastering technical skills used in mathematics BEFORE ever seeing any actual _ideas_ of which mathematics consists. – 2017-01-08

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$x^y=y^x$ : $x=y$ => o.k. $\enspace$(trivial like $x=x$)

$x>y$ => relation by parametrisation $x=(1+\frac{1}{t})^{t+1}$ and $y=(1+\frac{1}{t})^t$ with $t>0$.

For $y=2$ in the first case we have $x=2$ and in the second case we get $x=4$ using $t=1$ .

If $x<0$ and $y=2$ you can change to $x^2=(\frac{1}{2})^x$ with $x>0$ but maybe that's not what you are talking about.

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Your logic can only cover part of the solutions, but may not find out all of them.

If you are seeking real solutions only,

Take logarithm at both side

$$x^2=2^x$$ $$2\ln x=x\ln 2$$ $$\frac{\ln x}x=\frac{\ln2}2$$

Now consider the function $$f(x)=\frac{\ln x}x$$ $$f'(x)=\frac{1-\ln x}{x^2}$$

$f(x)$ is increasing in $(0,e)$ and decreasing in $(e,\infty)$.

Thus $f(x)=\frac{\ln2}2$ has exactly one solution at each of the region $(0,e)$ and $(e,\infty)$.

With observation we know the only positive real solutions are $x=2$ and $x=4$.

Edit:

As the comment and other answers suggested, there is also a negative solution given in terms of

the Lambert W function, which is defined as the inverse function of $f(z)=ze^z$.

We may use the identity

$$z=W(z)e^{W(z)}$$

Using this approach by @Aaron Maroja

$$ x=-\frac2{\ln2}W(\frac{\ln2} 2)\approx-0.766$$

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    ...and one can then use the Lambert function to display the other complex solutions. – 2017-01-08
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    no there is another solution – 2017-01-08
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    Here I assume the real solution only. I understand the concern though, but that already suggest OP's logic is incomplete. – 2017-01-08
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    As someone above noted, there is also a negative real solution (as can be seen graphically). – 2017-01-08
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    For negative real $x$ that $\log$ is wrong.. and there you will miss a solution. – 2017-01-08
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Short answer: no, not probably in the way you'd like. This is a transcendental equation and in general these must be solved graphically if you can't find solutions by inspection as in this case. (Even in this case, you haven't found all of the solutions... there are complex solutions).

However, if you wish, this particular transcendental equation has solutions in terms of the lambert W function $w(z)$, which is defined by the equation $z = w(z) e^{w(z)}$ (which itself cannot be solved by doing algebra).