Problem 41 - Precalculus

Question

Solve the logarithmic equations exactly. $$\log _{3}(2 x+1)=4$$

Step-by-Step Solution

Answer

The solution is \( x = 40 \).

Step 1: Understand the Logarithmic Equation

The given equation is \( \log_3(2x + 1) = 4 \). This means that, in base 3, the logarithm of \(2x + 1\) equals 4.

Step 2: Convert the Logarithm to Exponential Form

Using the property of logarithms that states if \( \log_b(a) = c \), then \( b^c = a \), we can rewrite the equation: \( 3^4 = 2x + 1 \).

Step 3: Calculate the Power of the Base

Compute \( 3^4 \). Since \( 3^4 = 81 \), rewrite the equation as \( 81 = 2x + 1 \).

Step 4: Isolate the Variable Term

Subtract 1 from both sides of the equation: \( 81 - 1 = 2x \), resulting in \( 80 = 2x \).

Step 5: Solve for \( x \)

Divide both sides of the equation by 2 to solve for \( x \): \( x = \frac{80}{2} = 40 \).

Key Concepts

exponential form

In mathematics, converting a logarithmic equation into exponential form is a key step to solving it. Essentially, an equation in logarithmic form such as \( \log_b(a) = c \) means that the number \( a \) is equal to the base \( b \) raised to the power of \( c \). In this way, the logarithm helps express a number as a power. To solve the problem \( \log_3(2x + 1) = 4 \), we use the property of logarithms to rewrite the equation in exponential form as \( 3^4 = 2x + 1 \). This transformation simplifies the equation, making it easier to solve by eliminating the logarithm and allowing us to perform algebraic operations directly on the expressions involved.

Performing the exponential operation \( 3^4 \), we calculate it as 81. So, the equation \( 3^4 = 2x + 1 \) becomes \( 81 = 2x + 1 \) when expressed in exponential form. The conversion to exponential form is crucial because it shifts our focus from finding a logarithmic result to solving a more straightforward algebraic equation.

properties of logarithms

Logarithms have several properties that make them a valuable tool for simplifying equations and solving problems. One essential property is the change from logarithmic to exponential form, which we have already seen. Here are some other important properties of logarithms that are often used in mathematical calculations:

• Logarithm of a Product: \( \log_b(m \cdot n) = \log_b(m) + \log_b(n) \)
• Logarithm of a Quotient: \( \log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \)
• Logarithm of a Power: \( \log_b(m^n) = n \cdot \log_b(m) \)

These properties are instrumental in rearranging and breaking down complex logarithmic expressions.

Understanding these properties not only helps in the algebraic manipulation of equations but also in converting logarithmic equations to exponential form or vice versa, as seen in the original exercise. Each of these properties stems from the relation of logarithms to exponents, allowing for a deeper comprehension of the connections between multiplication, division, and exponentiation through logarithmic expressions.

isolating variables

After converting the logarithmic equation into exponential form, the next crucial step is isolating the variable. Isolating a variable means rearranging the equation so that the variable is on one side of the equation by itself. This process often involves performing opposite operations to both sides of an equation.

In our example, once we have the exponential form \( 81 = 2x + 1 \), we need to isolate \( x \). Begin by subtracting 1 from both sides of the equation to get \( 80 = 2x \). Then, divide both sides by 2 to isolate \( x \) completely, resulting in \( x = 40 \).

Understanding how to isolate variables is an essential skill for solving algebraic equations. It allows one to find the value of the unknown quantity by systematically simplifying the equation. This method is widely used across different types of equations, not just logarithmic ones, making it a foundation of algebraic problem-solving.