Calculating logarithm without calculator

Question

I have a short question. How can I calculate a logarithm without a calculator? For example: Log base 8 of 4, log base 16 of 2.. Thanks.

Answer 1

$y = \log_{b} x $ is equivalent to $$ x = b^y$$
hence $y=\log_{8} 4 $ is equivalent to $$4= 8^y$$ $$ 2^2 = 2^{3y}$$ since you have same base then you can set the exponents to be equal. Therefore: $$ 2 =3y$$ so $$y=2/3 $$

Answer 2

for example we have $$\log_8 4=x$$ then we get $$8^x=4$$ and this is equivalent to $$2^{3x}=2^2$$ thus we get $$3x=2$$ or $$x=\frac{2}{3}$$ or write $$\log_{16} 2=x$$ then we get $$16^x=2$$ thus we get $$2^{4x}=2^1$$ and we get $$x=\frac{1}{4}$$

Answer 3

More general way to do it is as follows

1. Convert it to the natural logarithm for $N > 1$ $$\log_{10}(N) = \frac{\ln(N)}{\ln(10)}$$
2. Use the following property$$\ln(N)=-\ln\left(\frac1{N}\right)=-\ln\left(1 - \left(1-\frac1{N}\right)\right)$$
3. Use Taylor series for $\ln(N)$ up to a required precision $${\displaystyle {\begin{aligned}\ln(1-x)&=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\cdots &&{\text{ for }}|x|<1\\\ln(1+x)&=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots &&{\text{ for }}|x|<1\end{aligned}}}$$