The function #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)#?

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#g(x)=f(x)#

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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)#