Hello Jack,
First, we want to find the gradient of the line tangent to the given logarithmic curve where $x=1$. To do so, we must differentiate the curve with respect to $x$:
$$f'(x)=\frac{1}{x+1}$$
Now, we evaluate this for $x=1$:
$$f'(1)=\frac{1}{2}$$
Then, to find the possible gradients $m$ that make an angle of 45° with the tangent line at its point of tangency, we may equate the magnitude of the difference in the angles of inclination to $$\frac{\pi}{4}$$.
$$\left|\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right|=\frac{\pi}{4}$$
$$\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right)=\pm\frac{\pi}{4}$$
Taking the tangent of both sides, we find:
$$\tan\left(\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right)=\tan\left(\pm\frac{\pi}{4} \right)$$
Using the angle-difference identity for tangent on the left, and simplifying the right, we obtain:
$$\frac{m-\frac{1}{2}}{1+\frac{m}{2}}=\pm1$$
Now we may solve for $m$. Multiply through by $$2\left(1+\frac{m}{2} \right)$$
$$2m-1=\pm(2+m)$$
Square both sides, then arrange as the difference of squares:
$$(2m-1)^2-(2+m)^2=0$$
Apply the difference of squares formula:
$$(2m-1+2+m)(2m-1-2-m)=0$$
Combine like terms:
$$(3m+1)(m-3)=0$$
Hence the possible gradients are:
$$m=-\frac{1}{3},3$$
As we should expect, the product of the two gradients is -1 as the two lines would be perpendicular to one another.
