Intersection between $y=x$ and $y = \ln (x-2) + b$

Question

There is no context relating to this question or any previous information given. The question is just as follows:

The line $y=x$ is a tangent to the curve $\ln(x-2)+b$. Using calculus or otherwise, find the possible value(s) of $b$.

I wasn't sure how to approach this question though. The only thing I realized that there can only be one value for $b$, since $y=x$ is a tangent. What can I do now? I might just be overthinking a very simple question.

Answer 1

Let's differentiate the curve $y=\ln(x-2)+b$.

$$y'=\dfrac{1}{x-2}$$

If the curve $y=x$ is tangent to $y=\ln(x-2)+b$, both the curves will have same slope at the point of tangency. Since the slope of $y=x$ is a constant equal to 1, try equating $\dfrac{1}{x-2}$ to 1, to get the point of tangency.

$$\dfrac{1}{x-2}=1$$ $$x=3$$

So $x=3$ is the x coordinate of point of tangency. Since this point also lie on the curve $y=x$ the y coordinate is also $3$.

Substituting (3,3) in $y=\ln(x-2)+b$, we have $ b = 3 $.

Answer 2

Let $f(x)=x$ and $g(x)=\ln(x-2)+b.$

Then there is $x_0>2$ such that

$f(x_0)=g(x_0)$ and $f'(x_0)=g'(x_0).$

From the second equation we get $x_0=3$.

It is now your turn to determine $b$.

Answer 3

$y = x$ has slope 1.

Now, taking the derivative of $y = \ln(x - 2) + b$ and setting it equal to 1, we get

$$ y' = \frac{1}{x-2} = 1 $$

And solving for $x$, we have $1 = x - 2$; or rather, $x = 1$, which because of $y = x$, gives $y = 3$.

Finally, $ 3 = \ln(3 - 2) + b = 0 + b$; thus, $b = 3$.