Answer
To approximate $$0.99 \ln 0.98,$$
Notice that $0.99\approx 1$ and $0.98 \approx 1$ and we know that $0.99 \ln 0.98 \approx -0.02$.
We, therefore, let
$$z=f(x, y)=x \ln y$$
Then
$$
\begin{aligned}
d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\
&=\ln y d x+\frac{x}{y} d y
\end{aligned}
$$
To approximate $0.99 \ln 0.98 ,$ we let $x=1$, $d x=-0.01, y=1,$ and $d y=-0.02$ , so we have:
$$
\begin{aligned}
d z &=\ln (1) \cdot(-0.01)+\frac{1}{1}(-0.02) \\
&=-0.02 \\
f(0.99,0.98) &=f(1,1)+\Delta z \\
&\approx f(1,1)+d z \\
&=1 \cdot \ln (1)-0.02 \\
&\approx-0.02
\end{aligned}
$$
Thus, $0.99 \ln 0.98 \approx-0.02$
Using a calculator, $0.99 \ln 0.98 \approx-0.0200 .$
The error is approximately $0$ .
Work Step by Step
To approximate $$0.99 \ln 0.98,$$
Notice that $0.99\approx 1$ and $0.98 \approx 1$ and we know that $0.99 \ln 0.98 \approx -0.02$.
We, therefore, let
$$z=f(x, y)=x \ln y$$
Then
$$
\begin{aligned}
d z &=f_{x}(x, y) d x+f_{y}(x, y) d y \\
&=\ln y d x+\frac{x}{y} d y
\end{aligned}
$$
To approximate $0.99 \ln 0.98 ,$ we let $x=1$, $d x=-0.01, y=1,$ and $d y=-0.02$ , so we have:
$$
\begin{aligned}
d z &=\ln (1) \cdot(-0.01)+\frac{1}{1}(-0.02) \\
&=-0.02 \\
f(0.99,0.98) &=f(1,1)+\Delta z \\
&\approx f(1,1)+d z \\
&=1 \cdot \ln (1)-0.02 \\
&\approx-0.02
\end{aligned}
$$
Thus, $0.99 \ln 0.98 \approx-0.02$
Using a calculator, $0.99 \ln 0.98 \approx-0.0200 .$
The error is approximately $0$ .
