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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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1 Answer 1

up vote 2 down vote accepted

Don't get caught up in differentiating more than you need to. With a transcendental equation like yours, it's probably not going to give you an exact algorithmic solution.

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.

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