#e^x#+#lnx#+#2x#=#3# Find all possible values of x.?

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Answer 1 · 2018-07-19T15:06:23

There is only one solution #x=0.69#

Let

#f(x)=e^x+lnx+2x-3#

Then,

#f'(x)=e^x+1/x+2#

Apply Newton-Raphson Method

#x_(n+1)=x_n-f(x_n)/(f'(x_n))#

Let #x_0=1#

Then,

#x_1=1-f(1)/(f'(1))=0.816325#

#x_2=0.816325-f(0.816325)/(f'(0.816325))=0.69235#

#x_3=0.69235-f(0.692355)/(f'(0.692355))=0.69178#

graph{e^x+ln(x)+2x-3 [-4.933, 4.934, -2.465, 2.467]}