The idea of the first article is to rewrite any positive number $x$ as :
$$\tag{1}x=m\cdot 10^{\,e}$$ with $m$ the 'mantissa' : a real number between $1$ and $10$ (excluded)
and $\,e\,$ the 'exponent' (or "power of ten") which is a signed integer.
From this we have $\;\tag{2}\log_{10}(x)= e+\log_{10}(m)$
(to keep notations short I'll write $\,\log(x)\,$ for $\,\log_{10}(x)\,$ and use $\,\ln(x)$ for the 'natural' logarithm)
Since the mantissa is between $1$ and $10$ the idea is to memorize the first logarithms (here I'll use up to $5$ digits, you may use fewer or more digits if you prefer) :
$$
\begin{array} {ll}
m & \log\,m\\
1 & 0\\
2 & 0.30103\\
3 & 0.47712\\
4 & 0.60206\\
5 & 0.69897\\
6 & 0.77815\\
7 & 0.84510\\
8 & 0.90309\\
9 & 0.95424\\
\end{array}
$$
This seems to be much work but these values (except $\log(7)$) are not independent :
- $\log(2^n)=n\,\log(2)\;$ so that $\;\log(4)=2\log(2),\;\log(8)=3\log(2),\cdots$
- more generally $\;\log(a\cdot b)=\log(a)+\log(b)\;$ so that the table could be rewritten (using $\,1=\log(10)=\log(2)+\log(5)$) :
$$
\begin{array} {lll}
m & \log\,m\\
1 & 0\\
2 & 0.30103 &\log(2)\\
3 & 0.47712 &\log(3)\\
4 & 0.60206 &2\,\log(2)\\
5 & 0.69897 &1-\log(2)\\
6 & 0.77815 &\log(2)+\log(3)\\
7 & 0.84510 &\log(7)\\
8 & 0.90309 &3\,\log(2)\\
9 & 0.95424 &2\,\log(3)\\
\end{array}
$$
The table may thus be rebuilt with just three values (two if $\,\log(7)\,$ is omitted)!
One should memorize too $\;\ln(10) \approx 2.3026\,$ and its multiplicative inverse $\;\dfrac{1}{\ln(10)}=\dfrac{\ln(e)}{\ln(10)}=\log(e)\approx \dfrac 1{2.3026}\approx 0.4343$.
The classical expansion $\;\ln(1+x)\approx x\;$ will be rewritten as :
$$\tag{3}\log(1+x)=\frac{\ln(1+x)}{\ln(10)}\approx 0.4343\cdot x$$
(with $x$ replaced by $\;x-\dfrac{x^2}2+\dfrac{x^3}3-\cdots\,$ if needed)
Let's apply all this to recover the (unfortunately lost...) value of $\,\log(7)\,$ using only $\;\log(2),\log(3),\log(e)$. One method is to use $\;\log(7)=\dfrac{\log(7^2)}2\;$ and :
\begin{align}
\log(49)&=\log\left(50\;\left(1-\frac 1{50}\right)\right)\\
&=\log\left(\frac{100}2\right)+\log\left(1-\frac 1{50}\right)\\
&\approx\log(100)-\log(2)-0.4343\left(\frac 1{50}+\frac 1{2\cdot50^2}\right)\\
&\approx 2-\log(2)-\frac{0.8686}{100}\left(1+\frac 1{100}\right)\\
&\approx 2-0.30103-0.00877\\
&\approx 1.69020
\end{align}
Dividing by $2$ will indeed give the wished $0.84510$. Without the $\dfrac{x^2}2$ term we would get $\,0.84514$ : also not that bad!
$$-$$
Let's further practice and compute $\log(29012)$ as in your article : $\log(29012)=\log(2.9012\cdot 10^4)=4+\log(2.9012)$
In first approximation we have $\;\log(2.9012) \approx \log(3) \approx 0.477\;$ and deduce that $\log(29012)\approx 4+0.477 \approx 4.477$.
we have $\;2.9012\approx 3\cdot 0.9671 \approx 3\cdot (1-0.0329)\;$ so that
\begin{align}
\log(2.9012)&\approx \log(3)+\log(1-0.0329)\\
&\approx 0.47712-0.434\cdot 0.033\\
&\approx 0.47712-0.0143\\
&\approx 0.4628
\end{align}
and we got $\;\log(29012)\approx 4.4628\,$ not so far from the exact $4.4625776\cdots$
It is important to understand that the table of logarithms allows too the reverse computation that is to compute $10^{\,x}$.
Of course $10^{\,\log(2)}=2$ so that for example $10^{\,0.3}$ will be just a bit smaller than $2$.
For additional precision and for $x\ll 1$ let's write the useful $\displaystyle 10^x=e^{x\ln(10)}\approx 1+\ln(10)x\;$ or
$$\tag{4}10^{\,x}\approx 1+2.3026\cdot x$$
To compute $10^{\,x}$ decompose $x$ in its integer part $i$ and fractional part $f$
then $10^{\,i+f}=10^{\,i}\cdot 10^{\,f}$ : the mantissa of this result $10^{\,f}$ will be found using the table and $i$ will of course be the exponent.
After that all applications may follow : compute $a^b$ for any positive real $a$ and real $b$ using $\,\log(a^b)=b\,\log(a)\;$ so that
$$a^b=10^{\,b\log(a)}$$
Computing the $n$-th root of a positive real number will just be the special case $\;b=\dfrac 1n$.
Example $\sqrt[5]{1212}$ :
\begin{align}
1212&\approx 3\cdot 4\cdot 100\cdot 1.01\\
\log(1212)&\approx 2+\log(3)+\log(4)+0.4343/100\\
\dfrac{\log(1212)}{5}&\approx \frac{2+0.47712+0.60206+0.00434}5\\
\approx 0.6167
\end{align}
The answer (approximatively $10^{\,0.6167}$) will thus be a little over $4$.
More exactly $\,0.61670=0.60206+0.01464\,$ and since $\,10^{0.01464}\approx 1+2.3\cdot 0.0146 \approx 1.0336\;$ an approximate result will be $4\cdot 1.0336$ that is : $$\sqrt[5]{1212}\approx 4.134$$ while the exact result is $4.1371429\cdots$.
Many methods may be used to get more precision :
- from $1212=1200\cdot 1.01$ expand the logarithm (and/or the power of $10$) to second order as done for $\log(7)$
- compose (+ -) different exact values of logarithms to get values near the searched one (example $\,1.33\approx \dfrac 43$ and thus $\;\log(1.33)\approx \log(4)-\log(3)$)
- memorize values nearer to $1$ : $\log(1.1)\approx 0.041393,\ \log(1.2)=\log(\frac{3\cdot4}{10})\approx 0.07918$ and so on (you should nearly 'recognize' $\log(1.01)= 0.004321$... (see $\log(e)$) and won't need to memorize $\log(1.001)$)
- $\cdots$
Wishing you much fun discovering yourself other tricks,
Best Answer
You want to find the value of $n$ for which $100n^2 \approx 2^n$. Taking logs, we get $2\log n + 7 \approx n$, so $n \approx 7 + 2\log 7 \approx 13$. Of course, this is just an approximation.