Calculating logarithm without calculator
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Solution 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 $$y=2/3 $$
Solution 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 $$x=\frac{1}{4}$$
Solution 3
More general way to do it is as follows
- Convert it to the natural logarithm for $N > 1$ $$\log_{10}(N) = \frac{\ln(N)}{\ln(10)}$$
- Use the following property$$\ln(N)=-\ln\left(\frac1{N}\right)=-\ln\left(1 - \left(1-\frac1{N}\right)\right)$$
- Use Taylor series for $\ln(N)$ up to a required precision $${\displaystyle {\begin{aligned}\ln(1-x)&=-\sum _{n=1}^{\infty }{\frac {x^{n}}{n}}=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\cdots &&{\text{ for }}|x|<1\\\ln(1+x)&=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}=x-{\frac {x^{2}}{2}}+{\frac {x^{3}}{3}}-\cdots &&{\text{ for }}|x|<1\end{aligned}}}$$
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Author by
Barry
Updated on December 12, 2022Comments
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Barry 11 months
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.
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Henry over 6 yearsUse $\log_b(b^a)=a$ which translates to $\log_{c^n}(c^m)=\dfrac{m}{n}$
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CiaPan over 6 yearsIf you precede the logarithm name with a backslash:
\log
it will become a LaTeX symbol, rendered in upright font and looking like a function name: $\log$, instead of italic fontlog
→ $log$, which looks somewhat like a product of three varables $l\cdot o\cdot g$.