Problem 26 - Algebra 2

Question

Solve each equation. Check your solutions. \(\log _{4} a+\log _{4} 9=\log _{4} 27\)

Step-by-Step Solution

Answer

The solution is \(a = 3\).

Step 1: Apply the Product Rule for Logarithms

The equation is \(\log_{4} a + \log_{4} 9 = \log_{4} 27\). We can use the product rule of logarithms, which states \(\log_b m + \log_b n = \log_b (mn)\). Applying this to the left-hand side:\(\log_{4} (a \times 9) = \log_{4} 27\).

Step 2: Equate the Logarithmic Expressions

Now that both sides of the equation have the same base and are logarithms, we can equate the arguments. This gives \(a \times 9 = 27\).

Step 3: Solve the Equation for 'a'

Divide both sides of the equation by 9 to solve for \(a\): \(a \times 9 = 27\) becomes \(a = \frac{27}{9}\).

Step 4: Simplify the Solution

Simplify the fraction to find \(a\): \(a = \frac{27}{9} = 3\).

Step 5: Check the Solution

Substitute \(a = 3\) back into the original equation and check: \(\log_{4} 3 + \log_{4} 9 = \log_{4} 27\). Simplifying the left side using the product rule, we get \(\log_{4} (3 \times 9) = \log_{4} 27\), which simplifies to \(\log_{4} 27 = \log_{4} 27\). Since both sides are equal, \(a = 3\) is indeed a solution.

Key Concepts

Product Rule for Logarithms

The product rule for logarithms is a useful property that helps to simplify logarithmic expressions. This rule states that when you add two logarithms with the same base, you can combine them into a single logarithm by multiplying the arguments. Specifically, for logarithms with the base \(b\), the rule is expressed as:

• \(\log_b m + \log_b n = \log_b (mn)\)

Now let's look at how this rule applies to an equation. Suppose we have a problem like the one in the exercise: \(\log_{4} a + \log_{4} 9 = \log_{4} 27\). Using the product rule, we can combine the terms on the left side to form a single logarithmic expression:

• \(\log_{4} (a \times 9) = \log_{4} 27\)

At this point, the equation is simplified, with one logarithmic term on each side, allowing us to further progress in finding a solution. Remember, the product rule is only valid when the logarithms being added together have the same base, as is the case in the exercise. This allows the arguments to be multiplied, simplifying the expression for easier analysis.

Equating Arguments

Equating the arguments is a logical step that follows after utilizing the product rule for logarithms. Once both sides of the equation have the same base and are in the form of logarithms, you can equate their arguments. This is because if \(\log_b m = \log_b n\), it implies that \(m = n\), given that the logarithmic function is one-to-one.In the context of our exercise, after applying the product rule, the equation becomes:

• \(\log_{4} (a \times 9) = \log_{4} 27\)

The logarithms on each side are identical, allowing us to equate the arguments directly:

• \(a \times 9 = 27\)

Equating the arguments effectively reduces the complexity of the problem from dealing with logarithmic expressions to handling a simple algebraic equation. This step is crucial because it simplifies solving the original problem by working with the arguments directly.

Solving Equations

Once you have equated the arguments, solving the equation becomes straightforward. In this exercise, after equating the arguments, we have:

• \(a \times 9 = 27\)

To find the value of \(a\), you can isolate \(a\) by performing basic algebraic operations:

• Divide both sides by 9 to solve for \(a\): \(a = \frac{27}{9}\).

Simplifying the fraction gives us:

• \(a = 3\)

It’s important to check your solution by substituting back into the original equation. Substitute \(a = 3\) into \(\log_{4} a + \log_{4} 9 = \log_{4} 27\):

• See if \(\log_{4} 3 + \log_{4} 9 = \log_{4} 27\).

Simplify the left side again using the product rule: \(\log_{4} (3 \times 9) = \log_{4} 27\), confirming both sides are equal, hence verifying \(a = 3\) is indeed the correct solution. This step highlights the importance of solving equations using a systematic approach, including checking your work to ensure accuracy.