Find the equation of the straight line passing through the origin that is tangent to the curve y = ln(x).

Firstly, recognise the relevant equations. The two functions are y = mx and y = ln(x). As the straight line is a tangent, we know that at a certain point x0, the functions and their gradients are equal. Thus mx0 = ln(x0) [1], and by differentiating, m = 1/x0 [2]. [2] can be subbed into [1] to give 1 = ln(x0), and so by rearranging, x0 = e. This gives the point of intersection as (e, 1). By simply using gradient = rise/run, we can see that the equation of the desired line is given by y = e-1x.

PS
Answered by Paul S. Maths tutor

6251 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Complete the indefinite integral of 3x^2 + 4x -2/(x^2)


A ball is fired from a cannon at 20m/s at an angle of 56degrees to the horizontal. Calculate the horizontal distance the ball travels as well as its maximum height reached.


Given that (cos(x)^2 + 4 sin(x)^2)/(1-sin(x)^2) = 7, show that tan(x)^2 = 3/2


Demonstrate that (2^n)-1 is not a perfect square for any n>2, n ∈ N.


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences