How can I prove that $a^x$ < $a^{x + \Delta x}$, where $\Delta x > 0$, $a$ is a constant, and $a > 1$ (in my case, $a=2$)? I don't want to use a graph, of course.
Proof that $a^x$ < $a^{x + \Delta x}$
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inequality
exponentiation
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1For arbitrary constant, first, by proving all of math inconsistent. Try it with $0\lt a\lt 1$, $\Delta x\gt 0$. For $a=2$ and $\Delta x\gt 0$, take $a^{x+\Delta x}-a^x$; see if you get something positive. – 2010-11-24
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0Maybe use the fact that $f(x) = 2^x$ monotonic increasing and $2^{x+ \Delta x} = 2^{x}2^{\Delta x}$. – 2010-11-24
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0@Arturo OK, $a > 2$ (changed that a few seconds ago, but I got an error, and wasted a few seconds) $x > 1$ – 2010-11-24
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1No, not $a\gt 2$; you need $a\gt 1$ and $\Delta x\gt 0$; doesn't matter whether $x$ is positive or negative. – 2010-11-24
1 Answers
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It all depends on what you know. Do you know that if $a\gt 1$ and $b\gt 0$ then $a^b\gt 1$? If so, $2^{x+\Delta x}-2^x = 2^x(2^{\Delta x}-1)$ is a product of positive numbers.