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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?

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    $\begingroup$ If you are asked to do so, the result will be something simple that you can do using the laws of logs. In this case it is important that $128=2^7$. You should look for powers you know when doing these. $\endgroup$ – Ross Millikan Feb 13 '16 at 16:13
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    $\begingroup$ You could have written a more interesting example. What about $\ln(200.34)$ or $\log_{11}(4)$? $\endgroup$ – Von Neumann Feb 13 '16 at 23:16
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    $\begingroup$ @KimPeek The poster is likely a student who has recently studied logs and wishes to solve introductory exercises with exact answers. I do not believe the poster intended to approximate any logarithm. $\endgroup$ – zahbaz Feb 15 '16 at 1:19
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    $\begingroup$ slide rule works for me... $\endgroup$ – Joel Feb 15 '16 at 10:03
  • $\begingroup$ @Joel he said without a calculator :-) $\endgroup$ – Matt Gutting Feb 15 '16 at 12:17
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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 $$8^x = 8^{\frac{7}{3}} = (8^{\frac{1}{3}})^7 = 2^7 = 128$$

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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 $a>0,\ne1$

See Laws of Logarithms

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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!

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    $\begingroup$ The slide rule is a calculator. A mechanical calculator. $\endgroup$ – Matt Gutting Feb 15 '16 at 12:18
  • $\begingroup$ So how do you calculate logarithms with a slide rule? (No fair using the log scale or the loglog scales.) One way a slide rule can help is by providing accurate first approximations for the square roots that you need if you want to do the log by binary bracketing. A slide rule is also helpful in interpolating from a table. $\endgroup$ – richard1941 Mar 6 '18 at 17:16
  • $\begingroup$ Further, a slide rule facilitates interpolation when you are using a table. $\endgroup$ – richard1941 May 9 at 2:55
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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: $$ \log_a a = 1$$

$$ \log_y x^a= a \log_y x$$

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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, $$\log_{8}(128) = \log_{2^3}(2^7)=\log_{2^3}((2^3)^{\frac73})=\frac73$$ $$\log_{27}(2187) = \log_{3^3}(3^7)=\log_{3^3}((3^3)^{\frac73})=\frac73$$

$$\log_{36}(216) = \log_{6^2}(6^3)=\log_{6^2}((6^2)^{\frac32})=\frac32$$

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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.

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    $\begingroup$ The question was about logarithms, not square roots... $\endgroup$ – User that is not a user May 7 at 23:57
  • $\begingroup$ Could you explain your "and so"? Both times it comes out of nowhere. $\endgroup$ – hawkeye Sep 21 at 7:13
  • $\begingroup$ @hawkeye: Done. $\endgroup$ – Pieter Geerkens Sep 21 at 8:13
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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!

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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

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    $\begingroup$ This answer fails to give any information in how to obtain this result without a copy of this book. $\endgroup$ – User that is not a user May 7 at 23:44
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Try Ln(x) = Lim(n->inf)(n*(x^(1/n) -1). 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.

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    $\begingroup$ Knuth offers a method that uses SQUARING instead of taking the square root. This can be much easier, but unless you carry a lot of digits, the algorithm goes bad. Investigation of this is assigned as a 20 point problem. Assume your number is between 1 and the base. Square it. If the square is still less than the base, write a zero. Otherwise write a 1 and divide the square by the base. Repeat until you can't stand it any more. What you have is the log in binary. A good way to investigate this is to simulate it in a spreadsheet to see where things go wrong. $\endgroup$ – richard1941 Mar 6 '18 at 17:23

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