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.

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
$$
  \exp(0.9303) = \mathrm{e} \cdot \left( 1 - 0.0697 \right) = 2.71828 \cdot 0.9303 = 2.5288
$$
