Question

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)`?

`g(x)=f(x)`

Answer 1

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

Explanation:

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)`