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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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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. – Ross Millikan Feb 13 at 16:13
You could have written a more interesting example. What about $\ln(200.34)$ or $\log_{11}(4)$? – Beta Feb 13 at 23:16
@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. – zahbaz Feb 15 at 1:19
slide rule works for me... – Joel Feb 15 at 10:03
@Joel he said without a calculator :-) – Matt Gutting Feb 15 at 12:17
up vote 40 down vote accepted

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 $\log_a1\ne\log_a(-1)^2$


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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The slide rule is a calculator. A mechanical calculator. – Matt Gutting Feb 15 at 12:18

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 only works 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 $\frac{c}b$.

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:

$\ 7^2 = 49 \approx 48 = 3 * 16 = 3 * 4 ^ 2$

and so

$\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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This answer is additional to awesome answers already given, especially, that of N. F. Taussig.

Definition of 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 calculator. I will tell you a method that I use: since $\exp 3 \approx 20$, you can take $\log 20 \approx 3$. Hence, to calculate $\log n$ in practical applications, first calculate $\log_{20} n$, then multiply it by $3$. Since $20$ is an integer, it is easier to work with it. For example, if we need to calculate $\log 34627486221$. We do the following arithmetics: $$20^8 = 2^8 10^8 = 25600000000\\ \log 25600000000 \approx 8 \cdot 3 = 24\\ \log 34627486221 = \log 25600000000 + \log (34627486221 / 25600000000) \approx 24 + \log 1.35 \approx 24.35$$ in which the relative error is less than $1/297$.

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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