Big-O notation: Prove that $3^x$ is $O(3^x - 2^x)$

Question

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!

Answer 1

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.