Problem 60 - Precalculus

Question

Solve the logarithmic equations. Round your answers to three decimal places. $$\log \left(x^{2}+4\right)=2$$

Step-by-Step Solution

Answer

The solutions are \( x = 9.798 \) and \( x = -9.798 \).

Step 1: Convert Logarithmic Equation to Exponential Form

Given the equation \( \log (x^2 + 4) = 2 \), we first convert it to its exponential form. Remember that \( \log_b(a) = c \) is equivalent to \( a = b^c \). Here, the base is 10, so we write: \[ x^2 + 4 = 10^2 \] \[ x^2 + 4 = 100 \]

Step 2: Isolate the Term with the Variable

Subtract 4 from both sides to isolate the term containing \( x \): \[ x^2 = 100 - 4 \] \[ x^2 = 96 \]

Step 3: Solve for x

Take the square root of both sides to solve for \( x \). Remember to consider both the positive and negative square roots: \[ x = \sqrt{96} \] \[ x = 9.798 \] (rounded to three decimal places) \[ x = -9.798 \] (rounded to three decimal places)

Step 4: Verify Solutions in Original Equation

Substitute \( x = 9.798 \) and \( x = -9.798 \) back into the original logarithmic equation to validate them: For \( x = 9.798 \): \[ \log((9.798)^2 + 4) = \log(100) = 2 \] For \( x = -9.798 \): \[ \log((-9.798)^2 + 4) = \log(100) = 2 \] Both solutions satisfy the original equation.

Key Concepts

Exponential Form

Converting logarithmic equations into exponential form can significantly simplify solving them. When you see an equation like \( \log_b(a) = c \), this tells us that the base raised to the power of c equals a.
For example, in the exercise \( \log(x^2 + 4) = 2 \), we identify the base as 10, which is the presumed base for a logarithm when none is specified. This converts the equation to \( x^2 + 4 = 10^2 \).
Now, the equation reads as "\( x^2 + 4 \) is the result of 10 raised to the power of 2." Since \( 10^2 \) equals 100, we solve for the content inside the logarithm more easily.
This straightforward conversion to exponential form is a crucial step for finding solutions, particularly in equations where logarithms are used to compress exponential changes.

Square Root

After isolating the term \( x^2 = 96 \), the next step is finding the square root to solve for \( x \). The square root function has both positive and negative roots, reflecting that both \( (9.798)^2 \) and \( (-9.798)^2 \) equal 96.
The reason for considering both roots is that squaring either a positive or negative number yields the same positive result. Thus, \( x = \pm \sqrt{96} \).
In our example, calculating the square root of 96 gives us approximately 9.798. Make sure to address both the positive and negative possibilities in your solutions.
While the square root function is a straightforward arithmetic tool, taking both roots into account is essential when solving quadratic equations generated from logarithmic expressions.

Solution Verification

Once you find potential solutions by solving the transformed equation, verifying these solutions back in the original equation is vital. This ensures correctness and confirms that no steps were botched during simplification.
For the solutions \( x = 9.798 \) and \( x = -9.798 \), substitution back into \( \log((x)^2 + 4) = 2 \) verifies them.
It transforms to \( \log(100) = 2 \), which checks out because \( \log_{10}(100) = 2 \) is true.
Verification checks serve as confidence boosters and prevent overlooking mistakes, thereby affirming solution accuracy in math problems.