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

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Answer 1 · 2018-04-17T08:53:47

$x = - \frac{ln (100)}{0.005} \approx - 921.0340372$ $color(blue)(x=-ln(100)/0.005~~-921.0340372)$

By the laws of logarithms:

${log}_{a} (b^{c}) = c {log}_{a} (b)$ $log_a(b^c)=clog_a(b)$

${log}_{a} (a) = 1$ $log_a(a)=1$

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

Taking natural logarithms of both sides:

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

From above:

$- 0.005 x = ln (100)$ $-0.005x=ln(100)$

$x = - \frac{ln (100)}{0.005} \approx - 921.0340372$ $color(blue)(x=-ln(100)/0.005~~-921.0340372)$

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