Get $5$ by doing any operations with four $7$s How can one combine four sevens with elementary operations to get $5$? For example $$\dfrac{(7+7)\times7}{7}$$ (though that does not equal $5$). I am not able to do this. Can you solve it or prove that it's impossible?
 A: How about:
$$7 - \frac{7+7}{7} = 5$$
$$7 - \log_7 (7·7) = 5$$
$$7 - \frac{\ln (7·7)}{\ln 7} = 5$$
$$\left\lfloor \sqrt{\frac{7^7}{7!}} - 7\right\rfloor = 5$$
$$\lfloor 7\sin 777^\circ\rfloor = 5$$
$$\lfloor 7\cos 7^\circ\rfloor - \frac{7}{7} = 5$$
$$\lfloor 7\cos 7\rfloor = 5 \text{ using radians}$$
You can also use base $174$ and write:
$$\sqrt{\frac{77}{7·7}} = 5$$
That can also reduce the amount of sevens by one if you write:
$$\frac{\sqrt{77}}{7} = 5$$
A: With four 7's you can get any positive integer you want by just changing the number of squar roots in the following equation:
$$\frac{\ln\bigg{(}\frac{\ln(7)}{\ln(\sqrt{\sqrt{\sqrt{\sqrt{\sqrt{7}}}}})}\bigg{)}}{\ln\bigg{(}\frac{\ln(7)}{\ln(\sqrt{7})}\bigg{)}}=5$$
For example, you could get 35 by using 35 square roots. We aren't even really using the 7, you could equally use any other integer >1.
EDIT: Moved to only four 7's instead of five, based on Darth's suggestion.
A: Maybe:
$ \frac{((7^2-7)-7)}7=5$
A: OK, here's the ultimate: you can do it with a single $7$. It starts with the observation that when $n$ is smallish, $\ln n!$ is a little, but not much, greater than $n$. By repeatedly applying factorial, natural log and floor, we can get the value to creep up to around $5^2$.
$$\left\lfloor\sqrt{\ln\left(\left\lfloor\ln \left(\left\lfloor\ln \left(\left\lfloor\ln \left(7!\right)\right\rfloor!\right)\right\rfloor!\right)\right\rfloor!\right)}\right\rfloor = 5$$
I know you had four $7$s to use. But think of all the things you can do with the three $7$s you'll have left over!
A: Perhaps: $$7-\frac{7+7}{7}=7-2=5$$
A: How about
$$
\lfloor7-\ln7\rfloor=5
$$
I use only two $7$s.

$$
\lfloor7+7-\ln 7!\rfloor=5
$$
That's three $7$s.

But if you insist on using four $7$s
$$
\lfloor7-7+7-\ln 7\rfloor=5.
$$
A: Don't forget $$\frac{7! \mod 77}{7}$$ (and I haven't used any sneaky $2$s, either).
To get it down to three $7$s, you might try $$7 - \left\lceil .7 + .7 \right\rceil$$
You can also do it with only two $7$s: $$\left\lfloor{\sqrt 7 + \sqrt 7}\right\rfloor$$
(Can anyone do it with a single $7$?)
A: How about
$$
f(7, 7, 7, 7)
$$
where we define
$$
f(x, y, z, w) = 5
$$
for all $x, y, z, w$. (Clearly, the constant function on four parameters is an elementary operation.)
A: How about : $\Delta_{Q(\sqrt7)}/7+7/7$, $\Delta$ being the discriminant of a number field. 
A: 7-(7+7)/7
I started looking for a way to use the modulus operator, but this is too simple.
