How to solve logs with decimal bases and arguments no calculator

Question

For example, we have the following problem:

$$\log_{1.06}4.1 = x \implies 1.06^x = 4.1$$

How can you solve for an approximate $x$ without using a calculator?

Answer 1

Note

$$x = \log_{1.06} 4.1 = \frac{\ln 4.1}{\ln 1.06} = \frac{\ln 4+\ln (1+0.025)}{\ln (1+0.06)} \approx \frac{2\ln 2+0.025}{0.06}$$

Answer 2

I shall assume that you know the approximate value of a very few logarithms (in fact, you almost need to know for $2$, $3$, $5$, $7$ and $10$).

Suppose that you have $$x=\log_{a+\alpha}(b+\beta)=\frac{\log (b+\beta )}{\log (a+\alpha )}$$ and that $\alpha \ll a$ and $\beta \ll b$, then you can write $$\log(c+\gamma)=\log(c)+\log \left(1+\frac{\gamma }{c}\right)=\log(c)+\log \left(1+t\right) \qquad \text{where} \qquad t=\frac{\gamma }{c}$$ For more accuracy, you can write $$\log \left(1+t\right) =\frac{2 t}{2+t}$$ So, for your specific case $$x \sim \frac {\log(4)+\frac{2}{81}}{\log(1)+\frac{6}{103}}=\frac{103}{6} \left(\frac{2}{51}+2\log (2)\right)=24.4713$$ while the exact value should be $24.2151$

Answer 3

Rule of 72 says that if the rate times the time equals 72 you should double your money.

In this case, if you had money invested at 6% and you could double it every 12 years. If you have made 4.1 times on your money, you doubled it twice (and a touch more).

This gives a quick ballpark of $x = 24$

vs the $24.215$ I get with a calculator, not a bad estimate.