Question

find the equation of the line tangent to the graph of f at the indicated value of x. f(x)=\ln x^{6}; x = e^{2} y = (type an exact answer.)

Explanation:

Step1: Simplify the function

Using the property of logarithms $\ln x^n=n\ln x$, we have $f(x) = 6\ln x$.

Step2: Find the derivative of the function

The derivative of $\ln x$ is $\frac{1}{x}$, so by the constant - multiple rule, $f^\prime(x)=\frac{6}{x}$.

Step3: Find the slope of the tangent line at $x = e^2$

Substitute $x = e^2$ into $f^\prime(x)$. Then $m=f^\prime(e^2)=\frac{6}{e^2}$.

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

Substitute $x = e^2$ into $f(x)$. So $y=f(e^2)=6\ln(e^2)=6\times2 = 12$.

Step5: Use the point - slope form of a line

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(e^2,12)$ and $m=\frac{6}{e^2}$.
\[

$$\begin{align*} y-12&=\frac{6}{e^2}(x - e^2)\\ y-12&=\frac{6}{e^2}x-6\\ y&=\frac{6}{e^2}x + 6 \end{align*}$$

\]

Answer:

$y=\frac{6}{e^2}x + 6$