Question

What is the slope of the line tangent to the graph of `y=ln(x/2)` at x=4?

Answer 1

`y=1/4x-1+ln(2)`

Explanation:

First, find the point that the tangent line will intercept by plugging in `x=4`.

`y=ln(4/2)=ln(2)`

The point of tangency is `(4,ln(2))`.

Differentiating `y` will be simpler if we make use of the following logarithm rule:

`ln(a/b)=ln(a)-ln(b)`

Thus, we can redefine the function as

`y=ln(x)-ln(2)`

When we differentiate this, recall that `ln(2)` is a constant and can be ignored. Thus, the derivative of `y` is equivalent to just the derivative of `ln(x)`, which is `1"/"x`.

`y'=1/x`

The slope of the tangent line is equal to the value of the derivative when `x=4`, which is

`y'=1/4`

We know that tangent line has a slope `1"/"4` and passes through the point `(4,ln(2))`. These can be related as a line in point-slope form:

`y-ln(2)=1/4(x-4)`

Which can be rewritten as

`y=1/4x-1+ln(2)`

Graphed are the function and its tangent line:

graph{(y-ln(x/2))(y-ln(2)-(x-4)/4)=0 [-2.48, 13.32, -4.53, 3.37]}