Questions: Solve log4(x+3)+log4(2x+20)=2 x=

Transcript text: Solve \[ \begin{array}{l} \log _{4}(x+3)+\log _{4}(2 x+20)=2 \\ x= \end{array} \]

Solution

Solution Steps

To solve the equation \(\log_{4}(x+3) + \log_{4}(2x+20) = 2\), we can use the logarithm product rule, which states that \(\log_b(a) + \log_b(c) = \log_b(a \cdot c)\). This allows us to combine the logarithms into a single logarithm. Then, we can rewrite the equation in exponential form to solve for \(x\).

Step 1: Apply the Logarithm Product Rule

Combine the logarithms using the product rule:
\[ \log_{4}(x+3) + \log_{4}(2x+20) = \log_{4}((x+3)(2x+20)) \]

Step 2: Set the Equation Equal to 2

The equation becomes:
\[ \log_{4}((x+3)(2x+20)) = 2 \]

Step 3: Convert to Exponential Form

Rewrite the logarithmic equation in exponential form:
\[ (x+3)(2x+20) = 4^2 \]

Step 4: Simplify and Solve the Quadratic Equation

Simplify and solve:
\[ (x+3)(2x+20) = 16 \]
\[ 2x^2 + 20x + 6x + 60 = 16 \]
\[ 2x^2 + 26x + 60 = 16 \]
\[ 2x^2 + 26x + 44 = 0 \]
Divide the entire equation by 2:
\[ x^2 + 13x + 22 = 0 \]

Step 5: Factor the Quadratic Equation

Factor the quadratic:
\[ (x + 11)(x + 2) = 0 \]

Step 6: Solve for \(x\)

Set each factor to zero:
\[ x + 11 = 0 \quad \text{or} \quad x + 2 = 0 \]
\[ x = -11 \quad \text{or} \quad x = -2 \]

Step 7: Verify the Solutions

Check the solutions in the original equation. Only \(x = -2\) is valid because \(x = -11\) results in a negative argument for the logarithm.