Question

find the equation of the line tangent to the graph of f at the indicated value of x. f(x) = ln x^2 ; x = e^3 y = □ (type an exact answer.)

Explanation:

Step1: Simplify the function

Using the property $\ln x^{n}=n\ln x$, we have $f(x) = 2\ln x$.

Step2: Find the derivative of the function

The derivative of $\ln x$ is $\frac{1}{x}$, so $f^\prime(x)=\frac{2}{x}$.

Step3: Find the slope of the tangent - line at $x = e^{3}$

Substitute $x = e^{3}$ into $f^\prime(x)$. Then $m=f^\prime(e^{3})=\frac{2}{e^{3}}$.

Step4: Find the y - coordinate of the point of tangency

Substitute $x = e^{3}$ into $f(x)$. Since $f(x)=2\ln x$, then $y = f(e^{3})=2\ln(e^{3})=6$.

Step5: Use the point - slope form of a line

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(e^{3},6)$ and $m=\frac{2}{e^{3}}$.
$y - 6=\frac{2}{e^{3}}(x - e^{3})$.
Expand the right - hand side: $y - 6=\frac{2}{e^{3}}x-2$.
Solve for $y$: $y=\frac{2}{e^{3}}x + 4$.

Answer:

$y=\frac{2}{e^{3}}x + 4$