# no. of real roots of exponential equation in three questions

How Can i calculate no. of real roots of exponential equation in three questions

(1) $2^x = 1+x^2$

(2) $2^x+3^x+4^x = x^2$

(3) $3^x+4^x+5^x = 1+x^2$

My Try::

(1) Let $f(x) = 1+x^2-2^x$

now Diff. both side w.r. to $x$

$f^{'}(x) = 2x-2^x.\ln(2)$

and $f^{''}(x) = 2-2^x.\ln^2(2)$

Now for Max. or Min. $f^{'}(x) = 0$

means $2x = 2^x.\ln(2)$

Now How can I calculate after that

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For the first one, note that they are equal for $x = 0$. If $x < 0$, then $2^x$ has gotten smaller while $1 + x^2$ has gotten bigger, so they certainly aren't equal for $x < 0$.
Now, note that they are also equal at $x = 1$. For $x > 0$, at first $2^x$ is increasing faster. However, at some point in $(0, 1)$, $x^2$ begins increasing faster than $2^x$ and "catches up" at $x = 1$. Because $x^2$ is increasing faster, $x^2 > 2^x$ for $x$ near $1$. However, it's clear that eventually $2^x$ is larger eventually so at some point they will intersect again.
Here's a little bit more math. $\frac{d}{dx}[1 + x^2] = 2x$ and $\frac{d}{dx} 2^x = \ln 2 \cdot 2^x$. Hence, $1 + x^2$ is hardly increasing at all near $x = 0$. However, at $x = 1$, $1 + x^2$ is increasing at a rate of $2$ while $2^x$ is increasing at a rate of $2 \ln 2 < 2$. So at some point, $x^2 + 1$ began increasing faster. However, at $x = 4$, $2^x$ is increasing at a rate of $16 \ln 2$ which is much larger than $8$. There aren't going to be any more switches about which one is increasing faster. This is because a line can only intersect an exponential function at most twice, the line being $y = 2x$ and the exponential function being $y = \ln 2 \cdot 2^x$. We already know that there were two intersections, so there definitely aren't any more.