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#### MarkFL

Staff member
Here is the question:

LOGARITHMIC FUNCTION, MATHS QUESTION PLEASEE HELPP ME IM BEGGIN YOU!!!?

Consider the curve f(x)=ln(x+1). find the gradients of the possible tangents to f(x) which makes of 45 degrees with the tangent of f(x) at the point where x=1

I have posted a link there to this topic so the OP can see my work.

#### MarkFL

Staff member
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 $x$:

$$\displaystyle f'(x)=\frac{1}{x+1}$$

Now, we evaluate this for $x=1$:

$$\displaystyle f'(1)=\frac{1}{2}$$

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 $$\displaystyle \frac{\pi}{4}$$.

$$\displaystyle \left|\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right|=\frac{\pi}{4}$$

$$\displaystyle \tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right)=\pm\frac{\pi}{4}$$

Taking the tangent of both sides, we find:

$$\displaystyle \tan\left(\tan^{-1}(m)-\tan^{-1}\left(\frac{1}{2} \right) \right)=\tan\left(\pm\frac{\pi}{4} \right)$$

Using the angle-difference identity for tangent on the left, and simplifying the right, we obtain:

$$\displaystyle \frac{m-\frac{1}{2}}{1+\frac{m}{2}}=\pm1$$

Now we may solve for $m$. Multiply through by $$\displaystyle 2\left(1+\frac{m}{2} \right)$$

$$\displaystyle 2m-1=\pm(2+m)$$

Square both sides, then arrange as the difference of squares:

$$\displaystyle (2m-1)^2-(2+m)^2=0$$

Apply the difference of squares formula:

$$\displaystyle (2m-1+2+m)(2m-1-2-m)=0$$

Combine like terms:

$$\displaystyle (3m+1)(m-3)=0$$

$$\displaystyle m=-\frac{1}{3},3$$