Problem 39 - A Graphical Approach to Precalculus with Limits

Question

Solve each logarithmic equation. Express all solutions in exact form. Support your solutions by using a calculator. $$\log _{6}(2 x+4)=2$$

Step-by-Step Solution

Answer

The solution is \( x = 16 \).

Step 1: Understand the Equation

We are given a logarithmic equation \( \log_{6}(2x+4) = 2 \). This type of equation involves finding the value of \( x \) that makes the equation true.

Step 2: Convert Logarithm to Exponential Form

Recall that a logarithmic equation can be converted to exponential form. The equation \( \log_{b}(a) = c \) is equivalent to \( a = b^c \). Apply this to the given equation: \( 2x + 4 = 6^2 \).

Step 3: Simplify the Exponential Expression

Calculate \( 6^2 \). This equals 36. Therefore, the equation becomes \( 2x + 4 = 36 \).

Step 4: Solve for x

To isolate \( x \), subtract 4 from both sides of the equation: \( 2x = 36 - 4 \). This simplifies to \( 2x = 32 \).

Step 5: Final Solution for x

Divide both sides by 2 to solve for \( x \): \( x = \frac{32}{2} \). Thus, \( x = 16 \).

Step 6: Verify the Solution

Use a calculator to verify the solution. Substitute \( x = 16 \) back into the original logarithmic equation to check: \( \log_{6}(2(16) + 4) = \log_{6}(36) = 2 \) which matches the given equation.

Key Concepts

Exponential Form

The concept of logarithmic and exponential forms are closely tied together. Understanding this relationship makes solving logarithmic equations much easier. A logarithmic equation like \( \log_{b}(a) = c \) can be transformed into an exponential equation. This transformation is crucial because it changes the way we interpret the relationship. In exponential form, the base \( b \) raised to the power \( c \) equals \( a \).

• For example, converting \( \log_{6}(2x+4) = 2 \) into exponential form gives us \( 2x+4 = 6^2 \).
• Here, \( 6 \) is the base, and \( 2 \) is the exponent, resulting in \( 2x+4 = 36 \).

By understanding this conversion, we can shift our focus from logarithms to exponentiation, simplifying the problem into a standard algebraic equation that is typically more straightforward to solve.

Solving Equations

Once we've translated our logarithmic equation into an exponential form, it leads us to an algebraic equation that we can solve using basic operations. The process becomes systematic and easy to handle.

• First, we reconcile the equation, so \( 2x + 4 = 36 \) by calculating \( 6^2 \).
• Next, we solve for \( x \) by isolating it on one side of the equation.
• Subtract 4 from both sides: \( 2x = 36 - 4 \) simplifies to \( 2x = 32 \).
• Finally, divide both sides by 2 to find \( x = 16 \).

Throughout these steps, staying organized and understanding each transformation ensures accuracy. Remember, what we're doing here is essentially just simplifying until \( x \) is by itself. These steps handle the reasoning behind the calculations and pave the way to solve similar problems efficiently.

Verification of Solutions

Verifying your solution is a crucial step in problem-solving, ensuring that the answer properly satisfies the original equation. This confirmation step is often neglected but provides confidence in your mathematical reasoning. Calculators come in handy to verify solutions rapidly.

• In our exercise, substitute \( x = 16 \) back into the original logarithmic equation.
• Recalculate to check: \( \log_{6}(2(16) + 4) = \log_{6}(36) \).
• The calculation affirms the equation \( \log_{6}(36) = 2 \).

This match shows that the calculated \( x \) value is indeed correct. By practicing verification, you solidify your understanding of both the method and the solution's validity. It encourages a habit of thoroughness and precision in mathematics, which is key for success in more complex equations.