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
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