Logically deducing what's larger $0.8^{10}$ or $0.81^9$?

Question

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?

Are you sure you've written the two numbers properly? As it stands, I think it is clear which one is bigger. the challenging one would be $.8^9$ vs $.81^{10}$. — lulu, Aug 25 '17 at 01:04
Yes you are correct, I wrote it backwards. Sorry for that, thank you for pointing it out — user3002540, Aug 25 '17 at 01:29
Your first paragraph now has it opposite to both the title and second paragraph. You can edit the title as well as the body of the question. — Russell Borogove, Aug 25 '17 at 04:35

score 16 · Answer 1 · answered Aug 25 '17 at 01:08

16

$$(0.8)^{10}=(0.8)(0.8)^9<(0.8)^9<(0.81)^9$$

answered Aug 25 '17 at 01:08

score 3 · Answer 2 · answered Aug 25 '17 at 01:06

3

$$\frac{.8^{10}}{.81^9} = \left(\frac{80}{81}\right)^9(0.8)$$

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.

answered Aug 25 '17 at 01:06

G Tony Jacobs

31,218

IM SO STUPID. THIS is brilliant, simplistic. I'm so frustrated at myself, thank you. – user3002540 Aug 25 '17 at 01:22

score 2 · Answer 3 · answered Aug 25 '17 at 01:10

2

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)

answered Aug 25 '17 at 01:10

MT_

19,603
9
40
81

I'm sorry, I don't totally follow how you get from 8^9 > 9^8 = 8^(1/8) > 9^(1/9) – user3002540 Aug 25 '17 at 01:24
@user3002540 Take both sides to the power of $1/(8\cdot 9)$. – Michael L. Aug 25 '17 at 01:30

score 2 · Answer 4 · answered Aug 25 '17 at 01:22

2

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.

answered Aug 25 '17 at 01:22

Will Fisher

5,112

What do you mean by "the last inequality is fairly easy to show"? Please show us. – GregT Aug 25 '17 at 06:47
@GregT Just convert to fractions and do it by hand. Worst multiplication is $81\cdot 81$. – Will Fisher Aug 25 '17 at 16:42

Logically deducing what's larger $0.8^{10}$ or $0.81^9$?

4 Answers4