Problem 59 - Intermediate Algebra

Question

Solve each equation. $$ \log _{3} 4 x-\log _{3} 7=2 $$

Step-by-Step Solution

Answer

The solution is \( x = \frac{63}{4} \).

Step 1: Use Logarithmic Property

The given equation is \( \log_{3} 4x - \log_{3} 7 = 2 \). Use the property of logarithms that states: \( \log_{b} M - \log_{b} N = \log_{b} \left(\frac{M}{N}\right) \). Apply this property to combine the logarithms: \( \log_{3} \left( \frac{4x}{7} \right) = 2 \).

Step 2: Rewrite Logarithmic Equation into Exponential Form

To eliminate the logarithm, rewrite the logarithmic equation in its exponential form. This is done using the principle: \( \log_{b} A = C \implies A = b^C \). Apply this to get: \( \frac{4x}{7} = 3^2 \).

Step 3: Solve the Equation

Now compute \( 3^2 \) which equals 9. Substitute back to get \( \frac{4x}{7} = 9 \). Solve for \( x \) by isolating it: multiply both sides by 7 to get \( 4x = 63 \). Then, divide both sides by 4 to find \( x = \frac{63}{4} \).

Step 4: Simplify the Solution

Simplify \( x = \frac{63}{4} \) if possible. The fraction cannot be simplified further since 63 and 4 have no common factors other than 1. Therefore, the solution is \( x = \frac{63}{4} \).

Key Concepts

Logarithmic Properties

Logarithms have some amazing properties that help to simplify and solve equations. One key property is the subtraction rule for logarithms. This rule is so useful because it turns two separate logarithmic expressions into one. If you have an equation like \( \log_{b} M - \log_{b} N \), you can combine them by making it \( \log_{b} \left(\frac{M}{N}\right) \). In the current problem, \( \log_{3} 4x - \log_{3} 7 \) simplifies to \( \log_{3} \left(\frac{4x}{7}\right) \).

Why is this useful? It makes the equation simpler, with only one logarithm to deal with. From this point on, the equation is much easier to handle. Remember, using properties like this isn't just a trick, it's a crucial part of understanding logarithms and makes solving these complex-looking problems a lot easier.

Exponential Form

Switching a logarithmic equation to its exponential form sheds a new light on the problem. Essentially, you're transforming the equation into something more recognizable and, often, easier to solve.

For example, if you have \( \log_{b} A = C \), you can change it to \( A = b^C \). It's like flipping a switch on the equation. In our current problem, once we simplified using logarithmic properties, we had \( \log_{3} \left(\frac{4x}{7}\right) = 2 \).

By converting this into exponential form, it transforms to \( \frac{4x}{7} = 3^2 \). Now, instead of dealing with logarithms, you're working in the realm of basic algebra with an equation like \( 4x = 63 \).

• It helps make the problem more straightforward.
• You can more clearly see the steps needed to isolate the variable.

This little transformation trick is super handy in helping you solve logarithmic equations.

Algebraic Simplification

After converting to an exponential form, the next step is solving using simple algebra. Having an equation like \( \frac{4x}{7} = 9 \) is perfect for good old-fashioned algebra practice. To simplify, you need to solve for \( x \).

Here's a quick guide on how to do it:

• First, eliminate the fraction by multiplying every term by the denominator. In this case, multiply both sides by 7, giving you \( 4x = 63 \).
• Then, isolate \( x \) by dividing both sides by 4. This yields \( x = \frac{63}{4} \).

This straightforward process is often called isolation of the variable. Even though \( x = \frac{63}{4} \) looks a little complex, when you check it you'll see it can't be simplified further because there are no common factors between 63 and 4 other than 1.

With algebra, the key to simplifying equations is breaking them down into smaller steps, so you can handle each one smoothly. That's how algebra gives you the power to tackle even the trickiest parts of logarithmic equations!