Find a useful approximation for ln(x/y) when x is close to y, and give the next correction term.

Step-by-step solution

1. 1

   Turn the quotient into a small perturbation. Write $x=y+\Delta$ with $\Delta=x-y$ small compared with $y$. Then

   $\frac{x}{y}=1+\frac{\Delta}{y},\qquad\text{so}\qquad \ln\!\left(\frac{x}{y}\right)=\ln\!\left(1+u\right),\quad u=\frac{x-y}{y}.$

   The quantity $u$ is the relative change of $x$ over $y$. Recasting the problem in terms of $u$ is what makes a single series expansion do all the work.

2. 2

   Expand the logarithm about $u=0$. The Maclaurin series is

   $\ln(1+u)=u-\frac{u^{2}}{2}+\frac{u^{3}}{3}-\cdots,\qquad |u|<1.$

   The absence of a constant term reflects $\ln1=0$, and the leading coefficient $1$ is $\frac{d}{du}\ln(1+u)$ at $u=0$.

3. 3

   Read off the first-order approximation. Keeping only the leading term,

   $\boxed{\ln\!\left(\frac{x}{y}\right)\approx\frac{x-y}{y}}$

   This is the familiar statement that a log-difference approximates a percentage change: a $3\%$ rise gives a log change of about $0.03$.

4. 4

   Add the quadratic correction. Keeping one more term,

   $\ln\!\left(\frac{x}{y}\right)\approx\frac{x-y}{y}-\frac12\left(\frac{x-y}{y}\right)^{2}.$

   The correction is always negative for $x>y$, which says the first-order estimate systematically overstates the log of a rise — the logarithm is concave.

5. 5

   Note the alternative base point and quantify the error. Expanding about $x$ instead of $y$ gives $\ln\!\left(\frac{x}{y}\right)\approx\frac{x-y}{x}$, which understates the value; the two one-sided estimates bracket the truth, and dividing by the midpoint is more accurate than either. Numerically at $x=10.3$, $y=10$: the exact value is $0.0295588$, the first-order estimate is $0.03$ (error $1.5\%$), and the second-order estimate is $0.029550$ (error $0.03\%$) ✓.