Processing math: 100%

Rule of 72

$$(1+k\%)^{x}=2$$

$$x=\log_{1+k\%} 2=\frac{\ln2}{\ln(1+k\%)}$$

using L'Hospital's Rule:

$$\lim _{x\to c}{\frac {f(x)}{g(x)}}=\lim _{x\to c}{\frac {f'(x)}{g'(x)}}$$

so,

$$\lim _{x\to 0}{\frac {x} {\ln{1+x}}}=\lim _{x\to 0}{\frac {1}{\frac{1}{1+x}}}=\lim _{x\to0}{1+x}=1$$

so,

$$\frac{\ln2}{\ln(1+k\%)}\approx\frac{\ln2}{k\%}=\frac{100\times\ln2}{k}\approx\frac{69.3}{k}$$