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How can I prove that for any positive real number $C$, strictly greater than $1$, the function $$f(x)=C^x$$ is strictly increasing?

The definition of a strictly increasing function says that if $a<b$ then $f(a)<f(b)$.

Okay, suppose $a<b$. How to show that $C^a < C^b$?

My idea was to prove it by contradiction.

Let's say $C^a \ge C^b$. Taking $\log_c$ both sides (the fact the direction of inequality doesn't change follows from the fact that log is strictly increasing, so this proof doesn't change much actually):

$a\ge b$, a contradiciton, because we assumed $a<b$.

I'm actually wondering if it's possible to prove it without using derivatives.

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  • $\begingroup$ Show that the derivative is strictly positive $\endgroup$ – Miz Dec 16 '15 at 10:45
  • $\begingroup$ Suppose $a<b$, and notice that $C^b-C^a=C^a(C^{b-a}-1)$. It remains to show that $C^x>1$ if $x>0$. $\endgroup$ – Tom-Tom Dec 16 '15 at 10:50
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    $\begingroup$ @Tom-Tom actually it's the same problem. Even though I know that $C^x=1$ only for $x=0$, then to show that $C^x>1$ if $x>0$ requires that I know the function is strictly increasing. $\endgroup$ – user216094 Dec 16 '15 at 10:53
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    $\begingroup$ If you assume that $\log_C$ is strictly increasing, then it implies that its inverse, which is $C^x$, is also strictly increasing $\endgroup$ – glip-glop Dec 16 '15 at 10:57
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    $\begingroup$ What is your definition of $C^x$? $\endgroup$ – Crostul Dec 16 '15 at 11:16
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For the result to be true, you need $C>1$ (strict inequality)

$f'(x) = C^x \log(C) > 0$ since $C>1$

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I use your definition of $C^x$ as $$C^x= \exp ( x \log C)$$ where $\log C = \int_1^C \frac{1}{t} dt$ and $\exp$ is simply the inverse function of $\log$. It is clear that $\log$ is an increasing function, and so its inverse function $\exp$ is increasing as well.

Hence, for $x< y$ and $C > 1$ you have $\log C > 0$ so $$x \log C < y \log C$$ and applying $\exp$ you get the result.

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