Question

Express y as a function of \(x .\) The constant \(C\) is a positive number. \(\ln y=3 x+\ln C\)

Short answer

y = C e^{3x}

Step-by-step solution

- Understand the equation

The given equation is \(\backslashln y = 3x + \backslashln C\). We need to express \(y\) as a function of \(x\).

- Isolate ln y

Observe that \(\backslashln y\) is already isolated on the left side. The equation is already in a good form to exponentiate.

- Exponentiate both sides

To solve for \(y\), exponentiate both sides of the equation to eliminate the natural logarithm. \(\backslashexp{(\backslashln y)} = \backslashexp{(3x + \backslashln C)}\)

- Simplify using properties of exponents

Recall that \(\backslashexp{(\backslashln y)} = y\) and \(\backslashexp{(\backslashln C)} = C\). The equation becomes \ y = C \backslashexp{(3x)}\.

Key concepts

Natural Logarithm

Let's start by understanding what a natural logarithm is. A natural logarithm, denoted as \(\backslashln(...)\), is the logarithm to the base of the mathematical constant \(e\). The constant \(e\) is approximately equal to 2.71828. Natural logarithms are used to solve equations where the variable is in an exponent. For instance, if you have \(\ln y = 3x + \ln C \), you can use the property that will allow you to solve for \(y\). When you see \(\ln(...)\), it means the power to which you need to raise \(e\) to get the desired number.

Here are some key properties of natural logarithms:

• \(\ln 1 = 0\) because \(e^0 = 1\)
• \(\ln e = 1\) because \(e^1 = e\)
• \(\ln (ab) = \ln a + \ln b\) (Logarithm of a product)
• \(\ln (a^b) = b \ln a\) (Logarithm of a power)

Exponentiation

Exponentiation is the process of raising one number (the base) to the power of another (the exponent). To exponentiate means to take \(e\) to the power of both sides of the equation. If you have \(\ln y = 3x + \ln C\), exponentiating both sides would involve \(e^(\ln y)\) and \(e^{(3x + \ln C)}\). This operation helps eliminate the natural logarithm. Since \(e^{(\ln y)} = y\), we get simpler expressions.

Here are a few principles and properties of exponentiation:

• \(e^{(a + b)} = e^a \cdot e^b\) (Product of exponentials)
• \(e^{(ab)} = (e^a)^b\) (Power of an exponential)
• Exponentiation is the inverse operation of taking logarithms

Properties of Exponents

Understanding the properties of exponents is essential. These properties allow you to simplify and solve various mathematical expressions and equations. In the case of our example, using \(3x + \ln C\) in our exponentiation step, knowing these properties makes it easy to simplify:

Here are the most important properties:

• \(a^{m+n} = a^m \cdot a^n\) (Product of powers property)
• \(a^{mn} = (a^m)^n\) (Power of a power property)
• \((ab)^m = a^m \cdot b^m\) (Power of a product property)
• With \(e\), always remember that \(e^{\ln a} = a\)

Using these properties, exponentiating both sides of our equation \(\ln y = 3x + \ln C\) lets us simplify it to \(y = Ce^{3x}\). This is because \(e^{(3x + \ln C)} = e^{3x} \cdot e^{ \ln C}\), and since \(e^{\ln C} = C\), it further simplifies to \(y = C e^{3x}\).

Solving Equations

Finally, solving equations involves isolating the variable you want to solve for and performing operations that maintain the balance of the equation. In this exercise, we need to solve for \(y\). Given \(\ln y = 3x + \ln C\), the main strategy involves:

• Applying exponentiation on both sides to remove the natural logarithm
• Using properties of exponents to simplify

Steps to solve:
1. Exponentiate both sides: \(e^{\ln y} = e^{(3x + \ln C)}\)
2. Simplify: since \(e^{\ln y} = y\) and \(e^{\ln C} = C\), we get \(y = C e^{3x}\).

Follow these strategies whenever you encounter logarithmic equations. Start by isolating the logarithm if it’s not already isolated. Then exponentiate, and use exponent properties to simplify and solve the equation.