How do you solve #100e^(-0.6x)=20#?

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Answer 1 · 2016-11-16T14:15:58

#x =-0.767#

#100e^(-0.6x)=20#

#e^(-0.6x)=20/100=0.2#

Applying natural log on both sides

#log_ee^(-0.6x)=log_e0.2#

#-0.6xlog_ee=log_e0.2#

#-0.6x=2.303xx0.2#

#x=-0.767#