Root Calculation by Hand Is it possible to calculate and find the solution of $ \; \large{105^{1/5}} \; $ without using 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.
 A: 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. 
A: You can try using binomial theorem for real exponents.
You can write this as
$$ ((2.5)^5 + (105 - (2.5)^5))^{1/5} = 2.5 \left(1 + \frac{105 - (2.5)^5}{2.5^5}\right)^{1/5} = \frac{5}{2} \left(1 + \frac{47}{625}\right)^{1/5}$$
Taking first three terms of the binomial series 
$$(1+x)^r = 1 + rx + \frac{r(r-1)x^2}{2!} + \frac{r(r-1)(r-2)x^3}{3!} + \dots$$
using $r = \frac{1}{5}$ and $x = \frac{47}{625}$ gives us
$$ \frac{5}{2} \left(1 + \frac{47}{5*625} - \frac{4 * 47^2}{2*5^2*625^2}\right) = \frac{4954041}{1953125} \approx 2.5365$$
If you need a better approximation, you can include more terms.
All this can be done by hand using integer arithmetic, but is tedious.
A: I'm not exactly sure what you mean by 'without a calculator'.
You could try Newton's method to solve $f(x) = 0$, where $f(x) = x^5-105$. The Newton update is then $x_{n+1} = \frac{4}{5}x_n + \frac{1}{5} \frac{105}{{x_n}^4}$. This converges very quickly.
Of course, this involves computing the 4th power, and dividing...
A: Another way of doing this would be to use logarithm, just like Euler did:
$$
    105^{1/5} = \mathrm{e}^{\tfrac{1}{5} \log (105)} = \mathrm{e}^{\tfrac{1}{5} \log (3)}
 \cdot \mathrm{e}^{\tfrac{1}{5} \log (5)} \cdot \mathrm{e}^{\tfrac{1}{5} \log (7)}
$$
Use $$\log(3) = \log\left(\frac{2+1}{2-1}\right) = \log\left(1+\frac{1}{2}\right)-\log\left(1-\frac{1}{2}\right) = \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{2^{2k+1}} = 1 + \frac{1}{12} + \frac{1}{80} + \frac{1}{448} = 1.0.83333+0.0125 + 0.0022 = 1.09803$$
$$
  \log(5) = \log\frac{4+1}{4-1} + \log(3) = \log(3) + \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{4^{2k+1}} = \log(3) + \frac{1}{2} + \frac{1}{96} +\frac{1}{2560}
$$
$$
   \log(7) = \log\frac{8-1}{8+1} + 2 \log(3) = 2 \log(3) - \sum_{k=0}^\infty \frac{2}{2k+1} \cdot \frac{1}{8^{2k+1}} = 2 \cdot \log(3) - \frac{1}{4} - \frac{1}{768} 
$$
Thus
$$
  \frac{1}{5} \left( \log(3) + \log(5) + \log(7)\right) = \frac{4}{5} \log(3) + \frac{1}{5} \left( \frac{1}{2} - \frac{1}{4} + \frac{1}{96} - \frac{1}{768}  + \frac{1}{2560} \right) = \frac{4}{5} \log(3) + \frac{1993}{38400}= 0.9303 = 1-0.0697
$$
Now
$$
  \exp(0.9303) = \mathrm{e} \cdot \left( 1 - 0.0697 \right) = 2.71828 \cdot 0.9303 = 2.5288
$$
A: You can just do it by trial, but it gets tiring:  $2^5\lt 105 \lt 3^5$ so it is between $2$ and $3$.  You might then try $2.5^5 \approx 98$ so the true value is a bit higher and so on.
An alternate is to use the secant method.  If you start with $2^5=32, 3^5=243$, your next guess is $2+\frac {243-105}{243-32}=2.654$  Then $2.654^5=131.68$ and your next guess is $2.654-\frac {131.68-105}{131.68-32}=2.386$ and so on.  Also a lot of work.
Added:  if you work with RF engineers who are prone to use decibels, you can do this example easily.  $105^{0.2}=100^{0.2}\cdot 1.05^{0.2}=10^{0.4}\cdot 1.01=4 dB \cdot 1.01= (3 dB + 1 dB)1.01=2 \cdot 1.25 \cdot 1.01=2.525$, good to $\frac 12$%, where $1.05^{0.2}\approx 1.01$ comes from the binomial $(1+x)^n\approx 1+nx$ for $x \ll 1$
