Question

find the equation of the line tangent to the graph of f at the indicated value of x.
$f(x) = \ln x^9$ ; $x = e^4$
$y = \square$ (type an exact answer.)

Explanation:

Step1: Simplify the function

Using the property of logarithms \(\ln x^{n}=n\ln x\), so \(f(x)=\ln x^{9} = 9\ln x\).

Step2: Find the derivative of the function

The derivative of \(y = \ln x\) is \(y^\prime=\frac{1}{x}\). By the constant - multiple rule \((cf(x))^\prime = cf^\prime(x)\), the derivative of \(f(x)=9\ln x\) is \(f^\prime(x)=\frac{9}{x}\).

Step3: Evaluate the function and its derivative at \(x = e^{4}\)

• Evaluate \(f(x)\) at \(x = e^{4}\): \(f(e^{4})=9\ln(e^{4})\). Since \(\ln(e^{a})=a\), then \(f(e^{4})=9\times4 = 36\).
• Evaluate \(f^\prime(x)\) at \(x = e^{4}\): \(f^\prime(e^{4})=\frac{9}{e^{4}}\).

Step4: Use the point - slope form of a line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})\) is a point on the line and \(m\) is the slope. Here \(x_{1}=e^{4}\), \(y_{1}=36\), and \(m = \frac{9}{e^{4}}\).

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Answer:

\(y=\frac{9}{e^{4}}x + 27\)