Problem 336 - Precalculus

Question

Solve each equation for \(x.\) \(\log _{2}(7 x+6)=3\)

Step-by-Step Solution

Answer

\(x = \frac{2}{7}\)

Step 1: Understand the Equation

The given equation is in logarithmic form, \[ \log_{2}(7x + 6) = 3 \]The base of the logarithm is 2. The goal is to solve for \(x\) by first eliminating the logarithm.

Step 2: Convert to Exponential Form

To remove the logarithm, convert the equation into an exponential form using the logarithm definition: if \[ \log_b(y) = z, \text{ then } y = b^z.\]Applying this formula, we get: \[ 7x + 6 = 2^3. \]

Step 3: Simplify the Exponential Expression

Calculate the value of the expression on the right. \[ 2^3 = 8. \]So the equation simplifies to:\[ 7x + 6 = 8. \]

Step 4: Solve for x

Subtract 6 from both sides to solve for \(7x\): \[ 7x = 8 - 6 \]which simplifies to:\[ 7x = 2. \]Now divide both sides by 7:\[ x = \frac{2}{7}. \]

Key Concepts

Exponential Form

Understanding the relationship between logarithms and exponential functions can greatly simplify solving equations. Let's say you encounter a logarithmic equation like this: \[ \log_{b}(y) = z \] Here, \(b\) is the base, \(y\) is the number we're taking the logarithm of, and \(z\) is the resulting logarithmic value. In exponential terms, you can express this by raising the base \(b\) to the power \(z\) to retrieve \(y\): \[ y = b^z \] This transformation is what we call converting a logarithm to its exponential form. Doing this can often reveal the solution of an equation by stripping away the logarithm. For example, if you have \( \log_{2}(7x + 6) = 3 \), you can convert it to \[ 7x + 6 = 2^3 \] Thus transitioning from the logarithmic to the exponential view simplifies your problem, setting you on a direct path to find \(x\).

Logarithms

Logarithms might initially seem complex, but they are just another way to express exponents. Let's simplify. A logarithm answers the question: "To what exponent must the base be raised, to result in a particular number?" When you see \[ \log_{b}(y) = z \] it's asking: \(b\) raised to what power equals \(y\)? This means the logarithm is just representing the exponent itself.

Logarithms are useful because they allow you to handle equations involving exponentials, especially when trying to simplify or solve them. For example:

• They can transform multiplication into addition, making complex calculations manageable.
• By switching a logarithmic equation to an exponential one, they often simplify the solving process.

In the problem \( \log_{2}(7x + 6) = 3 \), the logarithm helps us convert the equation into an easier form to solve. Through understanding their behavior, logarithms become a powerful tool in algebra.

Solving Equations

Solving equations in mathematics includes isolating the unknown variable, typically denoted as \(x\), by performing operations that simplify and isolate \(x\) on one side of the equation. Here's a straightforward approach to solving the equation provided:
1. **Understand the equation**: Recognize the type of equation and what needs to be solved.
2. **Transform the form**, if necessary: Here, we turned \( \log_{2}(7x + 6) = 3 \) into its exponential counterpart \( 7x + 6 = 2^3 \).
3. **Simplify and calculate**: Solve for \(x\) by breaking down the complex operations into basic arithmetic. E.g., solve \(7x + 6 = 8\).

• Subtract 6 from both sides: \(7x = 2\)
• Divide by 7: \(x = \frac{2}{7}\)

This consistent sequence allows for a systematic way to tackle equations, ensuring that solving for \(x\) becomes straightforward and manageable.