ANSWER
[tex]\begin{gathered} L(x)=x-1 \\ \ln1.46=0.46 \end{gathered}[/tex]EXPLANATION
We want to find the linear approximation of the function at x = 1:
[tex]f(x)=\ln x[/tex]First, we have to find the equation of the tangent line to the function at x = 1:
[tex]\begin{gathered} f(1)=\ln(1) \\ f(1)=0 \end{gathered}[/tex]Now, find the derivative of the function at x = 1:
[tex]\begin{gathered} f^{\prime}(x)=\frac{1}{x} \\ \Rightarrow f^{\prime}(1)=\frac{1}{1} \\ f^{\prime}(1)=1 \end{gathered}[/tex]Now, find the equation of the line using the point-slope method:
[tex]y-y_1=m(x-x_1)[/tex]Therefore, we have:
[tex]\begin{gathered} y-0=1(x-1) \\ y=x-1 \end{gathered}[/tex]Hence, the linear approximation of f(x) at x = 1 is:
[tex]L(x)=x-1[/tex]To find the estimate of ln(1.46), substitute 1.46 for x in the equation above and simplify:
[tex]\begin{gathered} \ln1.46=L(1.46)=1.46-1 \\ \ln1.46=0.46 \end{gathered}[/tex]That is the answer.