Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
I would need to be able to compute logarithms without using a calculator, just on paper. The result should be a fraction so it is the most accurate. For example I have seen this in math class calculated by one of my class mates without the help of a calculator.
$$\log_8128 = \frac 73$$
How do you do this?
To evaluate $\log_8 128$, let $$\log_8 128 = x$$ Then by definition of the logarithm, $$8^x = 128$$ Since $8 = 2^3$ and $128 = 2^7$, we obtain \begin{align*} (2^3)^x & = 2^7\\ 2^{3x} & = 2^7 \end{align*} If two exponentials with the same base are equal, then their exponents must be equal. Hence, \begin{align*} 3x & = 7\\ x & = \frac{7}{3} \end{align*}
Check: If $x = \frac{7}{3}$, then $$8^x = 8^{\frac{7}{3}} = (8^{\frac{1}{3}})^7 = 2^7 = 128$$
Using $\log_xy=\dfrac{\log_ay}{\log_ax}$ and $\log(z^m)=m\log z$ where all the logarithms must remain defined unlike $\log_a1\ne\log_a(-1)^2$
$$\log_8{128}=\dfrac{\log_a(2^7)}{\log_a(2^3)}=\dfrac{7\log_a2}{3\log_a2}=?$$
Clearly, $\log_a2$ is non-zero finite for finite real $a>0,\ne1$
As you've seen, it can be a bunch of work to actually calculate them by hand. So, in the context of "no calculator", I'd like to point out that the slide rule was made almost exactly for this type of calculation!
Another way of doing this:
$$ 128= 2^7 = (2^3)^\frac{7}{3} = 8^\frac{7}{3}$$
$$ \log_8 128 = \log_8 (8)^\frac{7}{3} = \frac{7}{3}$$
Note the laws of logarithm used here: $$ \log_a a = 1$$
$$ \log_y x^a= a \log_y x$$
In general, this works only if the base of the logarithm is a power of some number. If it is, then write the base $b$ as $x^a$ for some integers $x,a$. Then try to write the argument of the log as $x^c$ for some integer $c$. Then the answer to the logarithm would be $\frac{c}a$.
For example, $$\log_{8}(128) = \log_{2^3}(2^7)=\log_{2^3}((2^3)^{\frac73})=\frac73$$ $$\log_{27}(2187) = \log_{3^3}(3^7)=\log_{3^3}((3^3)^{\frac73})=\frac73$$
$$\log_{36}(216) = \log_{6^2}(6^3)=\log_{6^2}((6^2)^{\frac32})=\frac32$$
All the answers have focused on the specific example you provide, but the numerical techniques involved readily apply to other examples as well. If you desire to obtain $\sqrt 3 $ note that:
$\ 7^2 = 49 \approx 48 = 3 * 16 = 3 * 4 ^ 2$
thus
$\ 3 \approx 7^2 / 4 ^ 2 = (7/4) ^ 2$
and taking the square root of both sides we yields
$\sqrt 3 \approx 7 / 4 = 1.75 $
Similarly for $\sqrt 2 $ note that $\ 10 ^ 2 = 100 \approx 2 * 49 = 2 * 7 ^ 2 $ and so $\sqrt 2 \approx 10 / 7 = 1.4 $
After some practice you will be able to get approximations within 1% very quickly, often in your head.
When pencil and paper are available one can often quickly double the precision through a single iteration of Newton's Method. For example:
$ \sqrt 2 / 1.4 \approx 1.42857 $ and so a better approximation is $ \sqrt 2 \approx (1.4 + 1.42857)/2 = 1.414285 $.
Repeating again gives $ \sqrt 2 \approx 1.41421356 $, which is as accurate as many hand calculators.
This answer is additional to awesome answers already given, especially, that of N. F. Taussig.
The definition of a logarithm in reals may help: $\log_b a$ is such a real number $c$ that satisfies $b^c = a$. For example, $\log_2 131072 = 17$ because $2^{17} = 131072$.
Also, you may want to be able to calculate natural logarithms without a calculator. I will tell you a method that I use: since $e^3 \approx 20$, you can take $\ln 20 \approx 3$. Hence, to calculate $\ln n$ in practical applications, first calculate $\log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it's easier to work with. For example, if we need to calculate $\ln 34 627 486 221$, we can do the following:
$$20^8 = 2^8 10^8 = 25 600 000 000\\ \log_{20} 25 600 000 000 \approx 8\\ \ln 25 600 000 000 \approx 8 \cdot 3 = 24\\ \ln 34 627 486 221 = \ln 25 600 000 000 + \ln (34 627 486 221 / 25 600 000 000) \approx 24 + \ln 1.35 \approx 24.35$$
The answer is only 0.13% off, which is very accurate.
Hope this helps!
In Apostol’s Calculus textbook, volume 1, the computational formula for the logarithm is developed. The specific example of $\log 2$ is given, obtaining the result $0.6921 < \log 2 < 0.6935$ “with very little effort," as Apostol remarks.
“You have no idea, how much poetry there is in the calculation of a table of logarithms!” -- Carl Friedrich Gauss
Try
$$\ln(x) = \lim_{n \to \infty} n \left( x^{1/n} - 1 \right)$$
If $n$ is a power of $2$, you get to take a lot of square roots. See the HHC 2018 proceedings for a paper on the computation of logarithms.
Generally, power series are efficient for natural logarithms of numbers near $1$. You can do things to get your number near $1$, such as multiplying by a power of ten or taking the square root, then adjusting the logarithm you get.
Meanwhile, memorize the number $0.4343$. That is the approximate logarithm of $e$. Use that to convert natural logs to base ten logs.
