How do you find the equation of the line tangent to #y=2^x# that passes through the point (1,0)?

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Answer 1 · 2016-06-04T15:10:56

#y = 2 e ln 2 (x-1)=3.76834(x-1)#, using 2 e ln 2 = 3.76834, nearly..

Equating natural logarithms,

#ln y =x ln 2#. Differentiating with respect to x,

#(y' )/ y =ln 2#. So, y'=y ln 2..

Now, at y = b, y' = b ln 2 and x = ln b/ln 2.

So, the equation of the line tangent at at y=b is

#y-b=b ln 2 (x-ln b/ln 2)#. If it passes through (1, 0),

#-b=b ln 2(1-ln b/ln 2)#. Solving for b,,

#ln b=1+ln 2#, and so, # b=e^(1+ln 2)=e^1 e^ln 2=2e#.

So, the slope of the tangent is b ln 2 = 2 e ln 2

Now, the line tangent through (1, 0) is given by

#y-0 = 2 e ln 2 (x-1)=3.76834(x-1)#

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