Solve the equation LNX = e + 1-x

Let f (x) = LNX + x-1-e, (x > 0),
f'(x)=1/x+1>0,
So f (x) increases monotonically, and f (E) = 0,
So the original equation has a unique solution, the solution is e

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1. 1. The number of solutions of the equation x ^ 2-x = LNX is 2. Given that the function f (x) = 4 ^ x + k * 2 ^ x + 1 has only one zero point, then the zero point is 1. The number of solutions of the equation x ^ 2-x = LNX is 2. Given that the function f (x) = 4 ^ x + k * (2 ^ x) + 1 has only one zero point, then the zero point is
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12. The derivative of X + LNX
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