Problem 22 - Intermediate Algebra

Question

For Problems \(15-22\), solve each logarithmic equation. $$ \ln (3 t-4)-\ln (t+1)=\ln 2 $$

Step-by-Step Solution

Answer

The solution is \( t = 6 \).

Step 1: Understand Logarithmic Properties

First, recall the property of logarithms that states \( \ln a - \ln b = \ln \left( \frac{a}{b} \right) \). We'll use this to combine the logarithms on the left-hand side of the equation.

Step 2: Combine Logarithms

Apply the log property from Step 1 to rewrite the equation: \( \ln \left( \frac{3t - 4}{t + 1} \right) = \ln 2 \).

Step 3: Eliminate the Logarithm

Since the natural logarithm is a one-to-one function, you can set the arguments equal to each other: \( \frac{3t - 4}{t + 1} = 2 \).

Step 4: Solve for t

Clear the fraction by multiplying both sides by \((t + 1)\) to get: \( 3t - 4 = 2(t + 1) \).

Step 5: Simplify and Solve the Equation

Distribute the 2 on the right-hand side: \( 3t - 4 = 2t + 2 \). Then, solve for \(t\) by subtracting \(2t\) from both sides: \( t - 4 = 2 \).

Step 6: Solve for t

Add 4 to both sides to solve for \(t\): \( t = 6 \).

Step 7: Verify the Solution

Plug \(t = 6\) back into the original equation to check: \( \ln (3(6)-4) - \ln(6+1) \) should equal \( \ln 2 \). Simplifying gives \( \ln 14 - \ln 7 = \ln 2 \), which is correct since \( \ln \left( \frac{14}{7} \right) = \ln 2 \).

Key Concepts

Natural Logarithm

The natural logarithm, often denoted as \( \ln \), is a logarithm with the base \( e \), where \( e \approx 2.71828 \). It's a special constant in mathematics that arises naturally in many areas, such as solving differential equations and modeling exponential growth.

When you see \( \ln(x) \), it means you're finding the power to which \( e \) must be raised to produce \( x \). For instance, \( \ln(e) = 1 \) because \( e^1 = e \).

Natural logarithms are prevalent in calculus, particularly with derivatives and integrals, given their unique mathematical properties. They are also used in scientific fields, such as biology and economics, to model continuous compound growth or decay. When solving equations involving natural logarithms, understanding how to manipulate them with their properties is key, which leads us to the next concept.

Properties of Logarithms

Logarithms have several useful properties that can simplify complex equations, particularly those involving multiplication, division, or exponentiation.

• Product Property: \( \ln(a \cdot b) = \ln a + \ln b \)
• Quotient Property: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \)
• Power Property: \( \ln(a^b) = b \cdot \ln a \)

In the original exercise, the Quotient Property was utilized. The equation \( \ln(3t-4) - \ln(t+1) \) was transformed using the formula \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \), simplifying the expression to \( \ln\left(\frac{3t-4}{t+1}\right) = \ln 2 \).

Applying these properties is crucial, as they allow the combination and simplification of logarithmic expressions, making solving equations straightforward. By transforming complex logarithmic terms, it becomes easier to isolate and solve for unknown variables.

Solving Equations

Solving logarithmic equations often involves a few strategic steps. First, you can use the properties of logarithms to condense or expand expressions, which help in isolating the variable.

To solve the equation \( \ln\left(\frac{3t-4}{t+1}\right) = \ln 2 \), the principle that logarithmic functions are one-to-one was applied. This means if \( \ln a = \ln b \), then \( a = b \).

Next, after equating the arguments, \( \frac{3t-4}{t+1} = 2 \), the equation was simplified by eliminating fractions, multiplying both sides by \( t + 1 \).

1. Distribute on both sides: \( 3t-4 = 2 \cdot (t+1) \).
2. Simplify: \( 3t-4 = 2t+2 \).
3. Solve for \( t \) by isolating terms: \( t = 6 \).

Finally, always verify your solution by substituting it back into the original equation. This ensures that no extraneous solutions are included. In this exercise, substituting \( t = 6 \) confirmed our solution was correct as \( \ln(14) - \ln(7) = \ln 2 \).

Following these steps helps tackle logarithmic equations systematically, leading to accurate solutions.