Is it possible to calculate and find the solution of 1051/5

withoutusing a calculator? Could someone show me how to do that, please?Well, when I use a Casio scientific calculator, I get this answer: 105^{1/5}\approx ” 2.536517482 “. With WolframAlpha, I can an even more accurate result.

**Answer**

I go back to the days BC (before calculators). We did have electricity, but you had to rub a cat’s fur to get it.

We also had slide rules, from which a 2 to 3 place answer could be found quickly, with no battery to go dead in the middle of an exam. Engineering students wore theirs in a belt holster. Unfortunately, slide rules were *expensive*, roughly the equivalent of two meals at a very good restaurant. For higher precision work, everyone had a book of tables.

My largish book of tables has the entry 021189 beside 105. This means that \log(105)=2.021189 (these are logarithms to the base 10, and of course the user supplies the 2). Divide by 5, which is trivial to do in one’s head (multiply by 2, shift the decimal point). We get 0.4042378.

Now use the tables backwards. The log entry for 2536 is 404149, and the entry for 2537 is 414320. Note that our target 0.4042378 is about halfway between these. We conclude that (105)^{1/5} is about 2.5365.

The table also has entries for “proportional parts,” to make interpolation faster. As for using the table backwards, that is not hard. Each page of the 27 page logarithms section has in a header the range of numbers, and the range of logarithms. The page I used for reverse lookup is headed “Logs .398\dots to .409\dots.”

There are other parts of the book of tables that deal with logarithms, 81 pages of logs of trigonometric functions (necessary for navigation, also for astronomy, where one really wants good accuracy). And of course there are natural logarithms, only 17 pages of these. And exponential and hyperbolic functions, plus a few odds and ends.

**Attribution***Source : Link , Question Author : Kerim Atasoy , Answer Author : André Nicolas*