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I was recently asked the following question.

Given $7^7$ and $8^7$, approximate $7.5^7$.

My idea was that we know $(x + .5)^7, (x - .5)^7$ where $x = 7.5$. Then, taking their average would over estimate by around $7.5^5$. Any other ideas?

EDIT: We are supposed to $90 \%$ sure of our answer which is a subjective term, so I assume our answer has to be pretty accurate.

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closed as off-topic by manshu, Gordon K, Fabich, Mike Earnest, Milo Brandt May 26 '16 at 15:36

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is off-topic as it appears to be a mathematics problem, as opposed to a mathematical puzzle. For more info, see "Are math-textbook-style problems on topic?" on meta." – manshu, Gordon K, Fabich, Mike Earnest, Milo Brandt
If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Do you know what level of accuracy the questioner wanted? $\endgroup$ – Gareth McCaughan May 26 '16 at 1:53
  • $\begingroup$ I'm impressed that one can average those numbers in their heads. How much arithmetic are we allowed to do before the solution is deemed to difficult to perform mentally? $\endgroup$ – Aurey May 26 '16 at 2:03
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    $\begingroup$ A closer estimate is to take the geometric mean instead: $\sqrt{7^78^7}=\sqrt{56}^7\approx\sqrt{56.25}^7=7.5^7$. $\endgroup$ – f'' May 26 '16 at 2:09
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    $\begingroup$ A "confidence interval" is a statistical term and doesn't make much sense in the context of this question. $\endgroup$ – f'' May 26 '16 at 2:11
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    $\begingroup$ Shouldn't this be migrated to Mathematics SE? $\endgroup$ – AMACB May 26 '16 at 15:49
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Here's a solution I feel isn't overly complicated, yet is pretty accurate (can be done on paper). Use the leading terms in a binomial expansion.

$(7.5)^7 = 7^7(1+0.5/7)^7 \approx 7^7(1+0.5/7*7+(0.5/7)^2*21)$

as $21$ is $7$ choose $2$.

This yields $7^7(1.5+0.25/7*3) \approx 7^7(1.5+0.1)$

which gives $1317668.8$, very close to the true value $\approx1334838.86$

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  • $\begingroup$ I guess if you apply the same reasoning from 8 and then perform an average on both values you should be able to be even nearer. $\endgroup$ – Walfrat May 26 '16 at 13:11
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Technically, I think your question is too broad. It reduces to: "How can you get a better approximation than by doing an arithmetic average without putting in much effort?" And I don't understand what "90% sure" means. If it means you have to be within 10%, then the solution is easy and it doesn't really qualify as a puzzle.

Nevertheless, since I have always found mental arithmetic really useful here's what I would do.

You know that 7.5 is .5 greater than 7. You know that $7 * 7 = 49$. Therefore, 7.5 is a little bit more than 7% larger than 7.

$7.5^7 > 7^7 * 1.07^7$

How much is $1.07^7$? Well, we know it's a bit more than $1 + (0.07 * 7)$ so it's pretty close to 1.5. This gives:

$7.5^7 > 7^7 * 1.5$.

Further, you know

$7.5^7 < (7^7 + 8^7) / 2$.

Now:

$7^7 = 823,543$
$8^7 = 2,097,152$

If you're really bad at mental arithmetic, you can round 823,543 (the minimum) down to 800,000 and $823,543 + 2,097,152$ (the maximum) up to 3,000,000 to get:

$7^7 * 1.5 < 7.5^7 < (7^7 + 8^7) / 2$
$800,000 * 1.5 < 7.5^7 < 3,000,000 / 2$
$1,200,000 < 7.5^7 < 1,500,000$

You know you're a fair amount bigger than the minimum and a fair amount less than the maximum, so take the midpoint: 1,350,000. This turns out to be within 2% of the correct answer. It takes a while to write and explain but you can do this in your head in a few seconds and be sure you are within a few percent of the correct figure. Absolutely sure that you are within 10% since subtracting 10% puts you just above the minimum and just below the maximum.

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  • $\begingroup$ You have $8^{8}$ where I think it should be $8^{7}$ $\endgroup$ – question_asker May 26 '16 at 12:55
  • $\begingroup$ The only problem I have with this is that you mentally compute $1.07^7$ and get $1.5$. How are you able to do this? $\endgroup$ – Trenin May 27 '16 at 14:05
  • $\begingroup$ @Trenin If the number gets 7% bigger 7 times, it's a little bit more than adding .07 seven times. Seven sevens are 49 so its a little more than 1.49. Using 1.5 makes the resulting math easier. $\endgroup$ – Hugh Meyers May 27 '16 at 18:11
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Assuming $7^7=a$ and $8^7=b$

I think we can approximate as below:

$${7.5}^7\times 1\approx \int_7^8x^7dx=\frac 18\left(8^8-7^8\right)=8^7-\frac 78 7^7=b-\frac 78 a$$

which gives $1376551.875$, very close to the true value $1334838.867$

This way is easy and fast.

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