# Fast Mental Calculation of $7.5^7$ [closed]

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.

• Do you know what level of accuracy the questioner wanted? May 26, 2016 at 1:53
• 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? May 26, 2016 at 2:03
• A closer estimate is to take the geometric mean instead: $\sqrt{7^78^7}=\sqrt{56}^7\approx\sqrt{56.25}^7=7.5^7$.
– f''
May 26, 2016 at 2:09
• A "confidence interval" is a statistical term and doesn't make much sense in the context of this question.
– f''
May 26, 2016 at 2:11
• Shouldn't this be migrated to Mathematics SE? May 26, 2016 at 15:49

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 $\approx1334838.86$

• 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. May 26, 2016 at 13:11

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.

• You have $8^{8}$ where I think it should be $8^{7}$ May 26, 2016 at 12:55
• 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? May 27, 2016 at 14:05
• @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. May 27, 2016 at 18:11

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.