Equation of the tangent to curve using logs?

Find the equation of the tangent to the curve y= ln(x) at the point (2, ln(2))
I just can't seem to work this one out! Any help is appreciated!

1 Answer
VNVDVI
Mar 21, 2018

#y=x/2-1+ln(2)#, or, written only in terms of logs, #y=x/2+ln(2/e).#

Explanation:

First, let's differentiate #y=ln(x).# The derivative of a function in terms of #x# gives us the rate of change (or slope) of that function at some value for #x.# Thus, we can use the derivative to find the slope of our tangent line at the given coordinates.

#y'=1/x# (The derivative of the natural logarithm function is #1/x#).

We want the slope at #(x_0,y_0)=(2,ln(2)),# so we evaluate our derivative at #x=2.# Note that the value for #y# plays absolutely no role here:

#y'(2)=1/2#

Now, we use the point-slope form of the line to determine the tangent line equation:

#y-y_0=m(x-x_0)# where #m# is our slope and #(x_0,y_0)# is some point on the line.

#m=1/2, # as calculated above.

#(x_0, y_0)=(2,ln(2))# as given to us by the problem. Thus, our tangent line equation becomes

#y-ln(2)=1/2(x-2)#

Solve for #y:#

#y-ln(2)=x/2-1#

#y=x/2-1+ln(2)#

We could further write #1# as #ln(e)# and continue simplifying. Totally optional.

#y=x/2-ln(e)+ln(2)#

#y=x/2+ln(2)-ln(e)#

#y=x/2+ln(2/e)# as #ln(a)-ln(b)=ln(a/b)#

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