Verified Answer :
- Substitute \(x=2\) into each equation: \(\log_2(5(2)+6) = \log_2(16) = 4\) (True), \(\log_2(16) = 4\) (True), \(\log_3(16) eq 3\) (False), \(\log_2(36) eq 6\) (False).
- Identify that the first and second equations satisfy the condition \(\text{LHS} = \text{RHS}\).
- Conclude that \(x=2\) is a solution only for these two.
- The equations with \(x=2\) as a solution are \(\boxed{\log_2(5x+6)=4 \text{ and } \log_x 16=4}\).
Explanation
- Problem Analysis and Strategy
To determine which of the provided equations have \(x=2\) as a solution, we will substitute \(x=2\) into each equation and check if the left-hand side (LHS) is equal to the right-hand side (RHS). The general form for a logarithm is \(\log_b(A) = C\), which is equivalent to \(b^C = A\), provided the base \(b > 0\) and \(b eq 1\).
- Evaluating the first equation
Consider the first equation: \(\log_2(5x+6) = 4\).
Substituting \(x=2\):
\(\text{LHS} = \log_2(5(2)+6) = \log_2(10+6) = \log_2(16)\).
Since \(2^4 = 16\), we have \(\log_2(16) = 4\).
Because \(\text{LHS} = 4\) and \(\text{RHS} = 4\), \(x=2\) is a solution to this equation.
- Evaluating the second equation
Consider the second equation: \(\log_x(16) = 4\).
Substituting \(x=2\):
\(\text{LHS} = \log_2(16)\).
Since \(2^4 = 16\), we have \(\log_2(16) = 4\).
Because \(\text{LHS} = 4\) and \(\text{RHS} = 4\), \(x=2\) is a solution to this equation.
- Evaluating the third equation
Consider the third equation: \(\log_3(6x+4) = 3\).
Substituting \(x=2\):
\(\text{LHS} = \log_3(6(2)+4) = \log_3(12+4) = \log_3(16)\).
Since \(3^3 = 27\), \(\log_3(16)\) is not equal to 3 (in fact, \(\log_3(16) \approx 2.52\)).
Because \(\text{LHS}
eq \text{RHS}\), \(x=2\) is not a solution to this equation.
- Evaluating the fourth equation
Consider the fourth equation: \(\log_x(36) = 6\).
Substituting \(x=2\):
\(\text{LHS} = \log_2(36)\).
Since \(2^6 = 64\), \(\log_2(36)\) is not equal to 6 (in fact, \(\log_2(36) \approx 5.17\)).
Because \(\text{LHS}
eq \text{RHS}\), \(x=2\) is not a solution to this equation.
- Conclusion
Based on our evaluation, the equations \(\log_2(5x+6)=4\) and \(\log_x 16=4\) are the ones that have \(x=2\) as a solution. Therefore, the valid options are the first and second equations. \(\boxed{\text{First and Second Equation}}\)
Examples
In engineering and computer science, verifying solutions to logarithmic equations is common when calculating signal decibel levels or algorithmic complexity. For example, if you are determining how many times a system needs to be divided to reach a certain threshold, a logarithmic equation models the process. If you find that a specific input (like \(x=2\)) satisfies the performance criteria defined by your logarithmic model, you can confidently proceed with that parameter in your system design.