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I have to show that $3^x$ is $O(3^x - 2^x)$. I'm just starting to learn the basics of Big-Oh notation. I'm thinking you have to take logarithms here, but am stuck on how to show this is true once I get $log(3^x-2^x)$ (although it makes intuitive sense to me that this is true). Would appreciate any help!

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    $\begingroup$ Please see math.meta.stackexchange.com/questions/5020 $\endgroup$ – Lord Shark the Unknown Nov 1 '18 at 7:01
  • $\begingroup$ Consider $$\frac{3^x}{3^x-2^x}.$$ $\endgroup$ – Lord Shark the Unknown Nov 1 '18 at 7:01
  • $\begingroup$ What is that limit? $\endgroup$ – Lord Shark the Unknown Nov 1 '18 at 7:05
  • $\begingroup$ How would I compute that limit? Taking the log of the numerator and denominator, I get $xlog(3)$ in the numerator but don't know what to do with $log(3^x-2^x)$ in the denominator. $\endgroup$ – bob Nov 1 '18 at 7:06
  • $\begingroup$ @bob You don't have to find the limit (there doesn't even have to be a limit for big-$O$ to make sense). You just have to show that that that fraction is (for large enough $x$) bounded both below and above by positive numbers. $\endgroup$ – Arthur Nov 1 '18 at 7:07
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Consider the fraction $\frac{3^x}{3^x-2^x}$. Clearly, for $x>0$, this fraction is larger than $1$ (the numerator is larger than the denominator), so it is bounded below. Not that this is actually necessary for $3^x=O(3^x-2^x)$.

For bounded above, which is necessary, consider $x>1$. Note that $2^x<2\cdot 3^{x-1}$. This gives $$ \frac{3^x}{3^x-2^x}<\frac{3^x}{3^x-2\cdot 3^{x-1}}=\frac3{3-2} $$ So the fraction is bounded above, and we're done.

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  • $\begingroup$ This is perfect, thank you for clarifying! $\endgroup$ – bob Nov 1 '18 at 7:23

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