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?

Solution 1:

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:

$$\frac{\sqrt{77}}{7} = 5$$

Solution 2:

Perhaps: $$7-\frac{7+7}{7}=7-2=5$$

Solution 3:

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 $7$?)

Solution 4:

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.