- Analyze the domain to ensure \(x > 0\).
- Simplify the logarithmic expression using the power and quotient rules to get \(\log_3(9x) = \log_3(x^2 + 4x + 4)\).
- Equate the arguments to form the quadratic equation \(x^2 - 5x + 4 = 0\).
- Factor the quadratic to find the solutions \(x = 1\) and \(x = 4\), matching option c.
- \(\boxed{c}\)
Explanation
- Analyzing the domain
We are given the logarithmic equation:
\[2 \log_3(6x) - \log_3(4x) = 2 \log_3(x+2)\]
First, we must determine the domain of the equation. Logarithmic functions \(\log_b(A)\) are defined only for \(A > 0\). Thus, we require:
- \(6x > 0 \implies x > 0\)
- \(4x > 0 \implies x > 0\)
- \(x + 2 > 0 \implies x > -2\)
Combining these, we find that the domain of the equation is \(x > 0\).
- Simplifying the equation using log rules
To solve for \(x\), we simplify the equation using logarithmic properties:
- Use the power rule \(\log_b(A^n) = n \log_b(A)\) to rewrite the terms:
\[2 \log_3(6x) = \log_3((6x)^2) = \log_3(36x^2)\]
\[2 \log_3(x+2) = \log_3((x+2)^2) = \log_3(x^2 + 4x + 4)\]
- Substitute these back into the equation:
\[\log_3(36x^2) - \log_3(4x) = \log_3(x^2 + 4x + 4)\]
- Apply the quotient rule \(\log_b(A) - \log_b(B) = \log_3(\frac{A}{B})\):
\[\log_3\left(\frac{36x^2}{4x}\right) = \log_3(x^2 + 4x + 4)\]
\[\log_3(9x) = \log_3(x^2 + 4x + 4)\]
- Solving the resulting quadratic equation
Since the logarithmic function is one-to-one, we can equate the arguments:
\[9x = x^2 + 4x + 4\]
Rearrange this into a standard quadratic equation:
\[x^2 - 5x + 4 = 0\]
We factor the quadratic expression:
\[(x - 1)(x - 4) = 0\]
This gives us two possible values for \(x\): \(x = 1\) or \(x = 4\). Since both values are within the domain \(x > 0\), both are valid solutions.
- Conclusion
The solutions to the equation are \(x = 1\) and \(x = 4\). Comparing this result with the given options, option 'c' matches our findings. Therefore, the true solutions are \(x=1\) and \(x=4\). The final answer is \(\boxed{c}\).
Examples
Logarithmic equations are used in various fields such as chemistry to calculate the pH of solutions, where \(\text{pH} = -\log[H^+]\), or in finance to determine the time needed for an investment to grow to a certain amount using compound interest formulas. By mastering the simplification of these equations, you gain a powerful tool for solving growth and decay problems that involve scaling factors or exponential relationships in the real world.