How do you solve $40 = 20 {(1.012)}^{x}$ $40 = 20(1.012)^x$ ?

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Answer 1 · 2018-07-03T20:46:49

$x = \frac{log 2}{log 1.012} = 58.10814961730409$ $x=log2/\log1.012=58.10814961730409$

Given that

$40 = 20 {(1.012)}^{x}$ $40=20(1.012)^x$

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

${(1.012)}^{x} = 2$ $(1.012)^x=2$

Taking logarithms on both the sides as follows

${log (1.012)}^{x} = log 2$ $log(1.012)^x=\log2$

$x log 1.012 = log 2$ $x\log1.012=log2$

$x = \frac{log 2}{log 1.012}$ $x=\log2/log1.012$

$= 58.10814961730409$ $=58.10814961730409$

How do you solve 40=20(1.012)x40 = 20(1.012)^x?