Question

A line with slope \(m\) passes through the origin and is tangent to \(y=\ln (2 x) .\) What is the value of \(m ?\)

Short answer

The value of m would be found by solving the equation m= \( \frac{ln(2x)}{x} \) which represents the condition that the tangency line also touches the curve at the tangency point.

Step-by-step solution

Expressing the function and finding its derivative using chain rule

The curve given is y = ln(2x). The derivative \(y'\) can be found using the chain rule. Chain rule implies \(\)derivative of ln(u) is \(1/u*\)derivative of u. Here, u = 2x. So differentiating w.r.t. x, we get \(y' = \frac{1}{2x}\times 2 = \frac{1}{x}\).

Equating the slope of the line to the derivative

The line is tangent to the curve, therefore the slope of the line (m) will equal the derivative at the point of tangency. As the line passes through the origin, the point of tangency on curve y=ln(2x) will be of the form (x,ln(2x)). Therefore, we equate m = \(y'\) = \(\frac{1}{x}\).

Solving for m

From the equation m = \(\frac{1}{x}\), the point of tangency also lies on the line. Thus it should satisfy the equation of the line y = mx. So, substituting y=ln(2x) into y=mx, we get this as m=\(\frac{ln(2x)}{x}\). Solving this equation would give the value of m.