- Identify the domain: the argument of the logarithm must be positive, so \(x > -11\).
- Simplify the right side of the equation: \(\log_5 25 = 2\), thus the equation becomes \(2 \log_5(x + 11) = 4\).
- Solve for the logarithm: \(\log_5(x + 11) = 2\), leading to \(x + 11 = 5^2 = 25\).
- Determine the value of \(x\): \(x = 25 - 11 = \boxed{14}\).
Explanation
- Domain and Problem Analysis
We are asked to solve the logarithmic equation:
\[2 \log_5(x + 11) = \log_5 25 + 2.\]
First, identify the domain of the expression. The logarithmic term \(\log_5(x + 11)\) is defined only if its argument is positive, so we must have:
\[x + 11 > 0 \implies x > -11.\]
This is our primary domain constraint. Any potential solution we find must satisfy this inequality to be valid.
- Simplifying the Right Side
Next, we simplify the right side of the equation. We know that \(25 = 5^2\), so the term \(\log_5 25\) becomes:
\[\log_5 25 = \log_5(5^2) = 2.\]
Substituting this back into the original equation, we get:
\[2 \log_5(x + 11) = 2 + 2,\]
\[2 \log_5(x + 11) = 4.\]
- Solving for x
Now, we solve for the logarithm. Divide both sides by 2:
\[\log_5(x + 11) = \frac{4}{2},\]
\[\log_5(x + 11) = 2.\]
To solve for \(x\), we convert the logarithmic equation to its exponential form. By the definition of the logarithm, \(\log_b(y) = k\) is equivalent to \(b^k = y\). Here, \(b = 5\), \(y = x + 11\), and \(k = 2\), so:
\[x + 11 = 5^2,\]
\[x + 11 = 25.\]
- Conclusion and Verification
Finally, solve for \(x\):
\[x = 25 - 11,\]
\[x = 14.\]
Since \(14 > -11\), the solution satisfies the domain constraint. We verify the solution:
Left side: \(2 \log_5(14 + 11) = 2 \log_5(25) = 2(2) = 4.\)
Right side: \(\log_5 25 + 2 = 2 + 2 = 4.\)
Both sides match, confirming that the solution is correct. The final answer is \(\boxed{14}\).
Examples
Logarithmic equations are used in various fields such as chemistry and finance. For instance, the pH level of a substance is determined by the formula \(\text{pH} = -\log_{10}[H^+]\), where \([H^+]\) is the concentration of hydrogen ions. Solving equations like these allows scientists to determine the precise concentrations needed to reach specific chemical balances or acidity levels in industrial manufacturing or environmental testing. Similarly, logarithms are used in compound interest calculations to determine the time required to reach a specific financial goal.