Chapter 4.4, Problem 31E

To determine

To calculate: The slope of line which is tangent to given curve.

Answer to Problem 31E

The slope is $\frac{2}{e}$ .

Explanation of Solution

Given information: The equation of curve is,

  $y=\ln \left(2x\right)$

And the line is passing through origin and is tangent to the curve.

The slope of two parallel lines are equal.

Formula used: $\left(slope\right):m_a=\frac{d{y}_a}{dx}$

Calculation: Slope of curve,

  $\begin{array}{l}{m}_1=\frac{dy}{dx}\\ m_1=\frac{d\left(\ln \left(2x\right)\right)}{dx}\\ m_1=\frac{1}{x}\\ \text{Calculate}\text{equation}\text{of}\text{line}\text{which}\text{is}\text{passing}\text{through}\left(0,0\right)\text{and}\left(x,y\right)\\ \left(y-0\right)=\frac{1}{x}\left(x-0\right)\\ y=1\\ \text{Equate}\text{this}\text{with}\text{original}\text{equation}\text{of}\text{curve,}\\ x=\frac{e}{2}\\ \text{So}\text{slope}\text{will}\text{be,}\\ m=\frac{2}{e}\end{array}$

The slope is $\frac{2}{e}$ .