Linear approximation to ln(x) at x = 1, then estimate ln(1.08)

Question

I know that the derivative of $\ln(x)$, or log of whatever base (x) = $(1/x)$ *the original function. If x is a more complicated expression, then the derivative would be $(x'/x)*f(x)$. If I knew the linear approximation, I would then plug .08 as the $dx$ term, and solve for whatever y is. However, how do I get the linear approximation for $ln(x)$ (log base $e$ of $x$)?

Answer 1

$y=\log(1+x)$, so $\frac{dy}{dx}=\frac{1}{1+x}$, the approximation is $log(1+x)=\frac{1}{1+x_0}$ $\Delta x$, where $\Delta x=x-x_0$ is the difference between the $x$ you want to calculate in the approximation and the x you used in the denominator of the derivative (that is, the point where you computed the slope). Choosing $x_0=0$ results in $log(1+x)=x$

Answer 2

The linear approximation in 'point-slope' form is an equation like $$(y-y_0) = m(x-x_0),$$ where $(x_0,y_0) = (1,\log(1)) = (1,0)$ is the point you're approximating around, and

$$m = \frac{d}{dx} \log(x)|_{x=1} = \frac{1}{x}|_{x=1} = 1$$ is the slope of the tangent.

You get your linear approximation $y-0 = 1(x-1)$, or $y = x-1$.

Plug in $1.08$ for $x$: $\log(1.08)$ is approximately $0.08$.