The function $f (x) = 4 e^{- x} + 2$ $f(x)=4e^-x+2$ , for $x inRR$ and $g(x)=2e^x-4$ , for $x inRR$ using the substitution $t=e^x$ , solve $g(x)=f(x)$ ?

Question

$g(x)=f(x)$

1473 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-12T15:56:38

$x=ln((3+sqrt17)/2)$

We have $f(x)=4e^(-x)+2$ and as $t=e^x$ , note that this means t>=0 $and we have$ f(t)=4/t+2#

and as $g(x)=2e^x-4$ , this becomes $g(t)=2t-4$

And $g(x)=f(x)$ translates to

$4/t+2=2t-4$

or $2t^2-4t=4+2t$

ot $2t^2-6t-4=0$

or $t^2-3t-2=0$

hence $t=(3+-sqrt(3^2-4*(-2)))/2=(3+-sqrt17)/2$

but as $(3-sqrt17)/2<0$ , we do not consider it

and $e^x=(3+sqrt17)/2$

and $x=ln((3+sqrt17)/2)$