How do you solve $e^(-0.005x) = 100$ ?

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Answer 1 · 2018-04-17T08:53:47

$color(blue)(x=-ln(100)/0.005~~-921.0340372)$

By the laws of logarithms:

$log_a(b^c)=clog_a(b)$

$log_a(a)=1$

$e^(-0,005x)=100$

Taking natural logarithms of both sides:

$-0.005xln(e)=ln(100)$

From above:

$-0.005x=ln(100)$

$color(blue)(x=-ln(100)/0.005~~-921.0340372)$

How do you solve e^(-0.005x) = 100 e^(-0.005x) = 100 ?