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

$:. x~=0.2197$

How do you solve 1.2e^(-5x)+2.6=31.2e^(-5x)+2.6=3?