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[1]:: $2^x = 1+x^2$

[2]:: $3^x+4^x+5^x = x^2$

My Trail for solution::

[1] Let $f(x) = 2^x-1-x^2$. Here Clearly $x = 0$ and $x = 1$ are solution of Given equation. Now We will find any other solution exists OR not.

So we use Derivative Test

$f^{'}(x) = 2^x \cdot \ln(2) - 2x$ and $f^{''}(x) = 2^x \cdot (\ln(2))^2-2$ and $f^{'''}(x) = 2^x \cdot (\ln(2))^3 > 0\forall x\in \mathbb{R}$

So Here $f^{'}(x)$ is Concave Upward.(Means Max. at Infty. but no idea about min.)

So $f^{'}(x) $ has at most $2$ real solution.

Now I did not understand it after that

Help Required.

Thanks