Logically deducing what's larger $0.8^{10}$ or $0.81^9$? Edit: I initially wrote the question backwards
I got a question recently, what's larger $0.8^9$ or $0.81^{10}$, don't have to calculate the exact number obviously, just figuring out which is larger. I'm not really sure how to proceed logically. I tried setting up as a divsion question. 
$0.8^{10}/0.81^9$, if the answer is greater than 1 I know the numerator is larger. I can rewrite this as : 
$$\frac{(8/10)^{10}}{(8.1/10)^9} = \frac{8^{10} *10^9}{8.1^9 *10^{10}}
= \frac{8^{10}}{8.1^9 * 10} $$
Beyond that, I don't see how to get the answer without using a calculator or doing some tedious math, I feel they're must be a logical approach. 
Edit: I figured it out by re-writing .81 as (.8*1.012) which then cancels out the numerator. Does anyone have a different method?
 A: $$\frac{.8^{10}}{.81^9} = \left(\frac{80}{81}\right)^9(0.8)$$
Both factors are smaller than $1$, so the product is clearly smaller than $1$. If there is a typo, and the numerator was supposed to be to a smaller power than the denominator, then this is still a decent approach.
A: Re lulu's comment, suppose the question is instead asking $.8^9$ versus $.81^{10}$. Then you can do something similar to get to
$$10 \cdot 8^9 \text{ vs. } 8.1^9$$
Multiply by $10^9$ we get
$$10^{10} \cdot 8^9 \text{ vs. } 9^{18}$$
Certainly $10^{10} > 9^{10}$, so it is sufficient to show that $8^9 > 9^8$, i.e. that $8^{1/8} > 9^{1/9}$, which is true since $x^{1/x}$ is decreasing on $x>e$. (This might be overkill, welcome to an easier solution)
A: We have seen many ways to show $0.8^{10}<0.81^9$, but as noted, showing $0.8^9>0.81^{10}$ is a little harder. A slightly simpler way than that of MCT is
$$\frac{0.8^{9}}{0.81^{10}}=\frac{1}{0.81}\cdot (1-1/81)^9\ge\frac{1}{0.81}\cdot(1-9/81)>1$$
where the last inequality is fairly easy to show. Also note that we use $(1-x)^n\ge 1-xn$ which is a direct consequence of the binomial theorem.
A: $$(0.8)^{10}=(0.8)(0.8)^9<(0.8)^9<(0.81)^9$$
