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Chapter 5: Q 91. (page 309)

Expressy as a function of x. The constant C is a positive number.
$\ln y = 3 x + \ln C$

Short Answer

Expert verified

$y = C e^{3 x}$

Step by step solution

01

Step 1. Given information.

Consider the given information.

$\ln y = 3 x + \ln C$

02

Step 2. Calculation.

Use the logarithm property to solve.

$\begin{array}{rcl} \ln y & = & 3 x + \ln C \\ \ln y - \ln C & = & 3 x \\ \ln \frac{y}{C} & = & 3 x \\ \frac{y}{C} & = & e^{3 x} \\ y & = & C e^{3 x} \end{array}$

03

Step 3. Conclusion.

The result is $y = C e^{3 x}$ .

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Most popular questions from this chapter

If $f (x) = a^{x}$ , show that $\frac{f (x + h) - f (x)}{h} = a^{x} . \frac{a^{h} - 1}{h} h \neq 0$ .

Solve given equation and verify your result using graphing utility.

${(e^{4})}^{x} . e^{x}$ ²=e12

Find $f (3)$ if $f (x) = - 4 x^{2} + 5 x$ .

Begin with the graph of $y = e^{x}$ and use transformation to graph the function. Determine the domain, range and horizontal asymptote of the function.

$f (x) = e^{x} - 1$

Use a calculator to compute the various values of the expression. Compare the values to e.

$2 + \frac{1}{1 + \frac{1}{2 + \frac{2}{3 + \frac{3}{4 + \frac{4}{e t c}}}}}$

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