Problem 33 - College Algebra

Question

Solve each logarithmic equation in Exercises \(27-44 .\) Be sure to reject any value of \(x\) that produces the logarithm of a negative number or the logarithm of \(0 .\) $$\log _{4}(3 x+2)=3$$

Step-by-Step Solution

Answer

The solution to the logarithmic equation \(\log _{4}(3x + 2) = 3\) is \(x = \frac{62}{3}\).

Step 1: Rewrite the Logarithmic Equation in Exponential Form

The given equation is \(\log _{4}(3x + 2) = 3\). In the first step, this logarithmic equation should be rewritten into its equivalent exponential form. Remember that the logarithmic equation \(\log _{b}(a) = c\) can be rewritten as \(b^c = a\). Hence, the equivalent exponential form of the equation becomes \(4^3 = 3x + 2\).

Step 2: Evaluate the Left Hand Side

Evaluate \(4^3\), which simplifies to \(64\). Now, the equation becomes \(64 = 3x + 2\).

Step 3: Solve for \(x\)

Subtract 2 from both sides to isolate the 3x term. This results in \(62 = 3x\). Finally, divide both sides by 3 to find the value of \(x\), which gives \(x = \frac{62}{3}\).

Step 4: Confirm That the Value Is Valid

Substitute \(x = \frac{62}{3}\) back into the original equation to ensure that it does not produce a logarithm of a negative number or zero. Substituting \(x = \frac{62}{3}\) into the equation \(\log _{4}(3x + 2) = 3\) results in a positive value, so \(x = \frac{62}{3}\) is a valid solution.

Key Concepts

Logarithm Properties

Logarithms are crucial tools in mathematics, particularly when dealing with exponential relationships. To understand how to solve logarithmic equations, we must familiarize ourselves with basic logarithm properties. One essential property is how to switch between logarithmic and exponential form—a concept employed when solving the provided exercise.

Another fundamental property is the logarithm of a product, expressed as \(\log_b(mn) = \log_b(m) + \log_b(n)\). This allows us to separate terms multiplied together into individual logarithms. Conversely, the logarithm of a quotient is expressed as \(\log_b(\frac{m}{n}) = \log_b(m) - \log_b(n)\), which simplifies division within a logarithm.

The power rule, \(\log_b(m^n) = n\cdot\log_b(m)\), is particularly helpful, as it allows us to move exponents in front of the log, simplifying calculations. Using these properties correctly is pivotal for solving more complex logarithmic equations accurately.

Exponential Form

An intuitive grasp of converting logarithmic statements to their exponential form is essential for solving logarithmic equations. The exponential form represents how a base raised to a particular power equals a number. When solving the exercise \(\log_4(3x + 2) = 3\), recall that the general form for a logarithmic equation \(\log_b(a) = c\) can be rewritten as \(b^c = a\), where \(b\) is the base, \(a\) is the result, and \(c\) is the exponent. Hence, translating \(\log_4(3x + 2) = 3\) to its exponential form gives us \(4^3 = 3x + 2\), rendering the previously opaque logarithmic equation into a more familiar algebraic one. This method is extremely versatile and applicable to all logarithms regardless of the base or complexity of the expression within the log.

Solving Algebraic Equations

When you've translated a logarithmic equation into its algebraic form, the next step is to solve for the unknown, much like the approach we took in the original exercise. In our case, the algebraic equation to solve was \(64 = 3x + 2\). Solving algebraic equations often involves a methodical process of isolation; an essential skill in algebra.

In general, our goal is to manipulate the equation to get the variable \(x\) by itself on one side. This typically involves performing inverse operations to remove any constants or coefficients linked with \(x\). As such, subtracting \(2\) from both sides, followed by dividing by \(3\), brought us to \(x = \frac{62}{3}\). It's also vital to ensure solutions are valid, especially for logarithms, as they're undefined for non-positive arguments. Therefore, checking that our \(x\) doesn't lead to a negative number or zero within the logarithm safeguards against invalid answers.

Mastering this method not only enables us to unravel logarithmic equations but also equips us with problem-solving strategies that span across various domains of mathematics.