Question

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

Answer 1

For `y=2^x`, we have `y'=2^x ln2`

The tangent line at `x=a`, has slope `m = 2^a ln2` and passes through the point `(a, 2^a)`.

So the equation of the tangent at the point `(a, 2^a)` is:

`y-2^a=(2^aln2)(x-a)`

We want `(1,0)` to be on the tangent line, so we want `(1,0)` to be a solution to the equation of the line.

That is:

We require: `0-2^a=(2^aln2)(1-a)`

Solve for `a = 1/ln2 +1 = lne/ln2 + 1 = log_2e +1`

So, `2^a = 2e`

`y-2^a=(2^aln2)(x-a)` becomes: `y-2e=(2eln2)(x-1/ln2 -1 )`

Which we can solve for `y` to get:

`y=(2eln2)x- 2eln2`