Question

Solve `e^x+logx=4` ?

Answer 1

`x = 1.3153148557141097`

Explanation:

A handy way to determine `x` is by using an iterative process like Newton-Raphson's

Calling

`f_k=e^(x_k)+logx_k-4` and
`df_k=e^(x_k)+1/x_k`

and

`x_(k+1)=x_k - f_k/(df_k)`

Beginning with `x_0 = 1` we obtain a convergent sequence

#((x_k,f_k),(1., -1.28172),(1.34471, 0.133239),(1.31562, 0.00137017),(1.31531, 1.468*10^-7),(1.31531, 1.77636*10^-15),(1.31531, 0.),(1.31531, 0.))#