Solving Log(1.66) without a calculator.

Question

I am taking a math class online in which I need to solve Algebra problems by hand without a calculator. I am getting caught up on how to solve log by hand. Whats a simple way to solve log(1.66)? we have to test via webcam with a piece of paper, and no calculator. Any advice?

Answer 1

If you don't need the exact solution, it can be done approximately. The series expansion for the natural log is $$ \ln(1+x)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} x^n = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \quad{\rm for}\quad \left|x\right| \leq 1,\quad $$ which seems to fit your purpose. For your test, you might want to try to memorize series expansions like this for trig functions too. Or you can derive them on the spot with Taylor's formula.

Answer 2

For logs you are basically in for Taylor series methods, supplemented by whatever values you have available in tables or memorized. You also have a tradeoff between how accurate you want to be and how much work you are willing to do. You didn't say what the base of the logs is. If it is base $10$, I know $\log_{10} 2\approx 0.30103$ and $\log_{10} 3 \approx 0.477$, so I would say $\log_{10} 1.66 \approx \log_{10} 5 - \log_{10} 3 \approx 1-0.301 - 0.477 = 0.222$. If they are natural logs, I know $\log 2 \approx 0.693$ so would say $\log 1.66 = \log 2 + \log (1-0.16) \approx 0.693 -0.16 + \frac {0.16^2}2-\frac {0.16^3}3$ where the next term is less than $0.001$ so I have three place accuracy. Dividing out the $2$ made the distance from $1$ smaller so the Taylor series converges more quickly.

Answer 3

There is one expansion which converges very fast

$$\log \left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$ Let us apply it for computing the logarithm of $\frac 53 =1.666667$; this would correspond to $x=\frac 14$. So $$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \frac{1}{5120}+\cdots\right)=\frac{3923}{7680}\approx 0.5108$$ Now, $$\frac {166}{100}=\frac 53\times\frac{249}{250}$$ So $$\log(1.66)=\log \left(\frac53\right)+\log \left(\frac{249}{250}\right)=\log \left(\frac53\right)+\log \left(1-\frac{1}{250}\right)\approx\log \left(\frac53\right)-\frac{1}{250}$$ $$\log(1.66)\approx 0.5108-0.0040=0.5068$$ while the exact value would be ... the same.

If you need less accuracy, just truncate the first part, that is to say $$\log \left(\frac53\right)=2\left(\frac 14+\frac{1}{192}+ \cdots\right)=\frac{49}{96}$$ $$\log(1.66)=\frac{49}{96}-\frac{1}{250}=\frac{6077}{12000}\approx 0.5064$$