Alex,

I would start by reminding myself of the definition of lag base 2.

s = log₂(t) if and only if t = 2^s         (1)

The x-intercept is the point on the curve where y = 0. From (1) is 0 = log₂(log₂x) then log₂(x) = 1 and then from (1) again x = 2. Thus the x-intercept is (2,0).

To find the slope of the tangent to the curve at this point you need to find the derivative of y = log₂(log₂x) with respect to x. This in going to involve the chain rule and the derivative of log₂x. Again using (1),

if s = log₂t then t = 2^s so ln(t) = ln(2^s) = s ln(2)

Differentiate both sides with respect to t to obtain

1/t = ds/dt ln(2)

and hence

if s = log₂t then ds/dt = 1/(t ln(2))

Use this fact and the chain rule to differentiate y = log₂(log₂x) and then evaluate the derivative at (2, 0).

Harley