Problem 34 - Algebra 2

Question

Solve each equation. Check your solutions. \(\log _{6}\left(a^{2}+2\right)+\log _{6} 2=2\)

Step-by-Step Solution

Answer

The solutions are \(a = 4\) and \(a = -4\).

Step 1: Apply the Product Rule of Logarithms

The given equation is \(\log _{6}(a^{2}+2)+\log _{6} 2=2\). Using the product rule of logarithms, which states \(\log_b(x) + \log_b(y) = \log_b(xy)\), we can combine the terms on the left side to get:\[\log _{6}((a^{2}+2) \cdot 2)=2\]. This simplifies to \(\log _6(2a^2 + 4) = 2\).

Step 2: Remove the Logarithm

To solve for \(a\), we need to remove the logarithm. Since \(\log_6(2a^2+4) = 2\), we can convert this logarithmic equation to its exponential form. The expression \(\log_b(x) = y\) implies that \(x = b^y\). Thus,\[2a^2 + 4 = 6^2\].

Step 3: Solve the Resulting Equation

We know \(6^2 = 36\). So the equation becomes \[2a^2 + 4 = 36\]. Subtract 4 from both sides to get: \[2a^2 = 32\]. Now, divide both sides by 2: \[a^2 = 16\].

Step 4: Find the Roots of the Equation

To find \(a\), take the square root of both sides: \(a = \pm \sqrt{16}\). This gives \(a = 4\) or \(a = -4\).

Step 5: Check the Solutions

Substitute \(a = 4\) back into the original equation: \[\log _{6}((4^2 + 2) \times 2) = \log _{6}(20 \times 2) = \log _{6}(40)\]. Since \(40 = 6^2\), the solution is valid for \(a = 4\). Now, substitute \(a = -4\): \[\log _{6}((-4)^2 + 2) + \log _{6}2) = \log _{6}(40)\]. Like before, this gives \(\log_6(40) = 2\). Hence, \(a = -4\) is also a valid solution.

Key Concepts

Product Rule of Logarithms

The product rule of logarithms is a key principle that allows you to combine two separate logarithms into one. If you have an equation with logs like \( \log_b(x) + \log_b(y) \), this rule tells us that you can combine them to become \( \log_b(xy) \). It simplifies calculations when dealing with multiple logs.
For instance, in our problem, we start with \( \log_6(a^2 + 2) + \log_6 2 = 2 \). Using the product rule, we turned this into \( \log_6((a^2 + 2) \cdot 2) = 2 \), resulting in \( \log_6(2a^2 + 4) \). By transforming the two logarithms into a single one, the equation becomes simpler and less cluttered, setting it up nicely for further steps.

Exponential Form

Once you've simplified the equation using logarithms, the next step is often to remove the logarithm entirely, which means converting it into exponential form. An equation \( \log_b(x) = y \) is equivalent to saying \( x = b^y \). This transformation is crucial for solving equations with unknown variables inside the log.

In the problem we're solving, we converted \( \log_6(2a^2 + 4) = 2 \) into its exponential form. By recognizing that \( 6^2 = 36 \), we can write \( 2a^2 + 4 = 36 \). Removing the logarithm paves the way for us to unfold the equation further and work directly with simpler mathematical expressions.

Solving Quadratic Equations

After converting our logarithmic equation into an exponential one, we end up with more straightforward forms like quadratic equations. In this particular exercise, the exponential form resulted in the equation \( 2a^2 + 4 = 36 \).

Here’s how we solve it:

• Subtract 4 from both sides: \( 2a^2 = 32 \).
• Divide every term by 2 to isolate \( a^2 \): \( a^2 = 16 \).
• Take the square root of both sides: \( a = \pm \sqrt{16} \), which gives us \( a = 4 \) or \( a = -4 \).

By these steps, we find potential solutions for \( a \) which we usually verify by checking.

Checking Solutions

Checking each solution ensures that they satisfy the original equation since sometimes extraneous solutions can appear, particularly with squared terms.

The process often involves substituting the values back into the original equation. Here:

• For \( a = 4 \): we check \( \log_6((4^2 + 2) \times 2) \), resulting in \( \log_6(40) \). Since \( 40 = 6^2 \), the solution \( a = 4 \) works.
• For \( a = -4 \): simplify \( \log_6((-4)^2 + 2) + \log_6 2 = \log_6(40) \). Again, this gives \( 40 = 6^2 \), confirming \( a = -4 \) is also valid.

Thoroughly checking the solutions ensures that all steps were correctly taken, verifying if they are true solutions to the initial problem.