# How do you solve 40 = 20(1.012)^x?

$x = \log \frac{2}{\setminus} \log 1.012 = 58.10814961730409$

#### Explanation:

Given that

$40 = 20 {\left(1.012\right)}^{x}$

${\left(1.012\right)}^{x} = \frac{40}{20}$

${\left(1.012\right)}^{x} = 2$

Taking logarithms on both the sides as follows

$\log {\left(1.012\right)}^{x} = \setminus \log 2$

$x \setminus \log 1.012 = \log 2$

$x = \setminus \log \frac{2}{\log} 1.012$

$= 58.10814961730409$