Question

find the equation of the line tangent to the graph of f at the indicated value of x.
f(x) = ln x^7; x = e^3
y = (type an exact answer.)

Explanation:

Step1: Simplify the function

Using the property $\ln x^n=n\ln x$, we have $f(x) = 7\ln x$.

Step2: Find the derivative

The derivative of $\ln x$ is $\frac{1}{x}$, so $f^\prime(x)=\frac{7}{x}$.

Step3: Find the slope of the tangent line

Substitute $x = e^{3}$ into $f^\prime(x)$. Then $m=f^\prime(e^{3})=\frac{7}{e^{3}}$.

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

Substitute $x = e^{3}$ into $f(x)$. So $y=f(e^{3})=7\ln(e^{3})=7\times3 = 21$.

Step5: Use the point - slope form

The point - slope form is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(e^{3},21)$ and $m=\frac{7}{e^{3}}$.
$y - 21=\frac{7}{e^{3}}(x - e^{3})$.
Expand to get $y - 21=\frac{7}{e^{3}}x-7$.
Then $y=\frac{7}{e^{3}}x + 14$.

Answer:

$y=\frac{7}{e^{3}}x + 14$