How do you solve #1.2e^(-5x)+2.6=3#?

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Answer 1 · 2016-07-20T18:11:59

#x=0.2ln3~=0.2197#

#1.2e^(-5x)+2.6=3#

#:. 1.2e^(-5x)=3-2.6=0.4#

#:. e^(-5x)=0.4/1.2=1/3#

#;. ln(e^(-5x))=ln(1/3)#

#:. -5xlne=ln(1/3)#

#:. -5x=ln(1/3)#

#:. x=-1/5ln(1/3)=-0.2(ln1-ln3)=-0.2(0-ln3)#

#:. x=0.2ln3#

Now, #0.2ln3=0.2(log_(10)3/log_(10)e)=0.2(0.4771/0.4343)~=0.2197#

#:. x~=0.2197#