Answer: \(\displaystyle \frac{3}{\ln2}\).

Answer: \(\displaystyle \frac{3}{\ln2}\).

At \(x=0\), the point \(A\) has

\(y=\frac{\ln2}{1}=\ln2.\)

So \(A=(0,\ln2)\).

Differentiate using the quotient rule:

At \(x=0\), this gives

\(\frac{dy}{dx}=-\ln2.\)

So the tangent gradient at \(A\) is \(-\ln2\), and the normal gradient is

\(\frac{1}{\ln2}.\)

The equation of the normal is therefore

\(y-\ln2=\frac{x}{\ln2}.\)

At point \(B\), \(y=0\). Hence

\(-\ln2=\frac{x}{\ln2},\)

so

\(x=-(\ln2)^2.\)

Thus \(B=(-(\ln2)^2,0)\).

Since \(C=(0,3\ln2)\), the gradient of \(BC\) is