Question

For what values of `k` will `(e^x)^k=x+1` generate exactly 2 solutions?

Answer 1

`k in RR, k>0, k!=1`

Explanation:

`(e^x)^k=e^(kx)`

What value(s) of `k` generate exactly 2 solutions of `x`:

`e^(kx)=x+1`

By playing around with the graph a bit, I think I can safely say that `k>0, k!=1` will generate two `x` solutions:

When `k<0`, `x` will have one solution:

graph{(y-e^(-x))(y-x-1)=0}

When `k=1`, `x` will also have just one solution:

graph{(y-e^(x))(y-x-1)=0}

With values of `0 < k < 1` (I used `1/2 and 1/5`):

graph{(y-e^(x/2))(y-e^(x/5))(y-x-1)=0[-5,20,-2,20]}

And with values of `k>1` (I used 2 and 10):

graph{(y-e^(2x))(y-e^(10x))(y-x-1)=0[-2,2,-2,5]}