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)#