Solve #e^x+logx=4# ?

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Answer 1 · 2017-04-13T17:54:21

#x = 1.3153148557141097#

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.))#