Find the linear approximation of f(x)=lnx at x=1 and use it to estimate ln1.38.
l(x)= ?
ln1.38 approximately = ?

 f(x) = ln(x) 
f'(x) = 1/x 

f(1) = 0 
f'(1) = 1 

Now we know the slope of the line, let's search for the costant term: 
1*x+a=f(x), for x=1: 

1*1+a=0 => a=-1 

linear approx: L(x) =1*x-1. 

L(1.38) = 1*1.38-1 = 0.38 ~= ln(1.38)

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The answer is poppy pants69