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

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Answer 1 · 2018-07-03T20:46:49

$x=log2/\log1.012=58.10814961730409$

Given that

$40=20(1.012)^x$

$(1.012)^x=40/20$

$(1.012)^x=2$

Taking logarithms on both the sides as follows

$log(1.012)^x=\log2$

$x\log1.012=log2$

$x=\log2/log1.012$

$=58.10814961730409$

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