Solve $e^{x} + log x = 4$ $e^x+logx=4$ ?

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Answer 1 · 2017-04-13T17:54:21

$x = 1.3153148557141097$ $x = 1.3153148557141097$

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

Calling

$f_{k} = e_{k}^{x_{k}} + {log x}_{k} - 4$ $f_k=e^(x_k)+logx_k-4$ and
$d f_{k} = e_{k}^{x_{k}} + \frac{1}{x_{k}}$ $df_k=e^(x_k)+1/x_k$

and

$x_{k + 1} = x_{k} - \frac{f_{k}}{d f_{k}}$ $x_(k+1)=x_k - f_k/(df_k)$

Beginning with $x_{0} = 1$ $x_0 = 1$ we obtain a convergent sequence

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

Solve e^x+logx=4ex+logx=4e^x+logx=4 ?