How can i quickly calculate an approximation of $\sqrt[5]{192}$ with just pen and paper How can I quickly calculate an approximation of $\sqrt[5]{192}$ with just pen and paper.
Because one time when I was in school, my calculator was broken, and they asked us on the exam to use calc to approximate $\sqrt[5]{192}$ to $1$ digit only.
 A: One should know some logarithms by heart. Among these is $\log 2\approx 0.3$ (which is quite good because $2^{10}=1024\approx 10^3$.
With this we have
$$\log 192\approx \log 200=2+\log 2\approx 2.3,$$
 hence $$\log\sqrt[5]{192}=\frac15\log192\approx 0.46$$
This looks quite close to $0.3+0.15=\log 2+\frac12\log 2$, so we have as approximation $$\sqrt[5]{192}\approx 2\sqrt 2\approx 2\cdot 1.4=2.8$$ because you of course also know this one-decimal approximation of $\sqrt 2$ :)
A: Doesn't get you any closer than Bill Dubuque's answer, but the technique generalizes to other problems.
Integer guess with one iteration of Newton's method:
$2^5 = 32$
$3^5 = 243$
$3$ Is closer so let $x_0 = 3$
Newton's method has: $$x_{n+1}=x_n - \frac{f(x_n)}{f'(x_n)}$$
With
$f(x) = x^5 - 192$
and
$f'(x) = 5x^4$
Gives one iteration as:
$$x_1 = x_0 - \frac{x_0^5 - 192}{5x_0^4} = 3 - \frac{3^5 - 192}{5 \times 3^4} = 3 - \frac{243 - 192}{5 \times 81} = 3 - \frac{51}{405} \approx 3 - \frac{50}{400} \approx 2.875$$
The calculations involved in iterating a second time more or less preclude the use of pen and paper, but if you're desperate, you do get a good payoff: Newton's method roughly doubles the number of correct digits each iteration.
A: Maybe you know that $2^5=32$.  It isn't too many multiplies to find that $3^5=243$, especially if you know that $3^4=9^2=81$.  So it is a little less than $3$.  Now $\frac{192}{243} \approx \frac {200}{1.04 \cdot 243} \approx \frac {200}{253} \approx 0.8$  Using the binomial theorem, $(1-x)^5 \approx 1-5x$, so $(1-\frac {0.2}5)^5=0.96^5 \approx 0.8$  Our approximation is $0.96 \cdot 3 \approx 2.88$  In fact, $2.88^5 \approx 198$ so we are a little off, but not badly.  The correct value is about $2.862$ so we are less than $1\%$ off
A: There are already some better methods here but I couldn't resist trying this out. If you like calculating these sorts of things you may have memorized that $\ln(2) \approx .7$ and $\ln(10)\approx 2.3$.
$$ \ln(192) \approx \ln(190) = \ln(200-10) = \ln(200) + \ln(1-1/20) = \ln(2)+2\ln(10)-\ln(1-1/20) \approx .7+2(2.3) - .05 \approx 5.3$$
Now we divide by $5$ to take the 5th root,
$$ \frac{5.3}{5} = 1+\frac{.3}{5} = 1+{.6}{10} = 1.06 \approx 1.05$$
Exponentiating our result we get,
$$ \sqrt[5]{192} \approx e^{1.05} = e^{1} e^{.05} \approx 2.7(1.05) = 2.7+\frac{2.7\cdot \frac{10}{2}}{100} = 2.7 + \frac{13.5}{100} = 2.7+.135=2.835 \approx 2.8$$
A: $(\overbrace{3^5\!-\!51}^{\large 192})^{1/5}\! = 3\left[1\!-\dfrac{51}{3^5}\right]^{\large 1/\color{#c00}5}\!\!\! = 3\left[1\!-\dfrac{51}{\color{#c00}5\cdot 3^5}+\,\cdots\,\right] \approx 3-\frac{50}{400} \approx 2.875\ $ (error $= +0.013)$
A: First of all, $\root{5}\of {192} = 2\root{5}\of{6}.$  Note that $1.5^5 > 7.5$ and and $\sqrt{2}^5 = 4\sqrt{2} = 5.6+.$  so you are between 2.8 and 3.0.  
A: $2^5 = 32$
$3^5 = 243$
$2 \le \sqrt[5]{192} \le 3$
For a little more accuracy, $192$ is pretty close to $243$. You can guess $2.8 \le \sqrt[5]{192} \le 3$.
A: For convenience, I'll denote $x = \sqrt[5]{192}$.
We know that $x^5 = 192$.  $2^5 = 32$, and $3^5 = 243$.  So, $2^5<x^5<3^5$, from which we conclude that $2<x<3$.
From here, you could guess that $x$ will be closer to $3$ than to $2$ because of how close $x^5$ is to $192$.  In order to check, you could verify that $x^5$ is bigger than $2.5^5$.
In fact, we may state
$$
2.5^5 = (2 + 0.5)^5 \approx 2^5 + 5\cdot(0.5) + 10 \cdot (0.5)^2\\
= 32 + 2.5 + 2.5 = 37
$$
Since $37$ is much less than $192$, we can assume $2.5^5$ is safely less than $192$, so that $x$ is closer to $3$.
A: Start with 
$$192=6\cdot2^5\Longrightarrow\sqrt[5]{192}=2\cdot6^{1/5}$$ 
then note that
$$6^2=36\approx32=2^5\Longrightarrow 6^{1/5}\approx2^{1/2}\approx1.4$$
to get $\sqrt[5]{192}\approx2.8$.
