Solution:
Given the function below
[tex]f(x)=\ln x,\text{ }at\text{ }x=1[/tex]
To find the linear approximation, the formula is
[tex]L(x)=f(a)+f^{\prime}(a)(x-a)[/tex]
Where
[tex]\begin{gathered} Where,\text{ }a=1 \\ f(1)=\ln1=0 \\ f^{\prime}(x)=\frac{1}{x} \\ f^{\prime}(1)=\frac{1}{1}=1 \end{gathered}[/tex]
Substitute into the linear approximation formula above
[tex]\begin{gathered} L(x)=f(a)+f^{\prime}(a)(x-a) \\ L(x)=0+1(x-1) \\ L(x)=0+x-1 \\ L(x)=x-1 \end{gathered}[/tex]
Hence, the linear approximation is
[tex]L(x)=x-1[/tex]
To estimate
[tex]\ln(1.31)[/tex]
Substitute 1.31 for x into the deduction above,
[tex]\begin{gathered} L(x)=x-1 \\ L(1.31)=1.31-1=0.31 \\ L(1.31)=0.31 \end{gathered}[/tex]
Hence, the answer is
[tex]\ln(1.31)\approx0.31[/tex]
