Resolved: "How do you find the equation of the..."

Frauffshiesiaf6s

Answered question

2022-02-10

How do you find the equation of the line tangent to

y = 2^{x}

that passes through the point (1,0)?

Keenan Mora

Beginner2022-02-11Added 11 answers

For

y = 2^{x}

, we have

y^{'} = 2^{x} \ln 2

The tangent line at x=a, has slope

m = 2^{a} \ln 2

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^{a} \ln 2) (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^{a} \ln 2) (1 - a)

Solve for

a = \frac{1}{\ln 2} + 1 = \frac{\ln e}{\ln 2} + 1 = \log_{2} e + 1

So,

2^{a} = 2 e

y - 2^{a} = (2^{a} \ln 2) (x - a)

becomes:

y - 2 e = (2 e \ln 2) (x - \frac{1}{\ln 2} - 1)

Which we can solve for y to get:

y = (2 e \ln 2) x - 2 e \ln 2

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?