Verified Answer :

  • Analyze each equation by testing \(x = -6\):
  • The first equation \(\log_x 36 = 2\) fails because the base \(x = -6\) is negative.
  • The second equation \(\log_3(2x - 9) = 3\) fails because the argument \(2(-6) - 9 = -21\) is negative.
  • The third equation \(\log_3 216 = x\) fails because \(\log_3 216 \approx 4.86 eq -6\).
  • The fourth equation \(\log_3(-2x - 3) = 2\) succeeds because the argument \(-2(-6) - 3 = 9\) is positive, and \(\log_3 9 = 2\) holds true.
  • Final Answer: \(\boxed{\log_3(-2x - 3) = 2}\)

Explanation

  1. Analysis of the Problem

We are tasked with determining which of the four given logarithmic equations has \(x = -6\) as a valid solution. A logarithmic equation of the form \(\log_b y = z\) is defined only if the base \(b > 0\) and \(b eq 1\), and the argument \(y > 0\). We will evaluate each equation one by one by substituting \(x = -6\) and checking these conditions.

  1. Evaluating the first equation: \(\log_x 36 = 2\)

For the first equation, \(\log_x 36 = 2\), substituting \(x = -6\) gives a base of \(-6\). By the definition of logarithms, the base must be greater than zero and not equal to 1. Since \(-6 < 0\), this equation is not defined for \(x = -6\).

  1. Evaluating the second equation: \(\log_3(2x - 9) = 3\)

For the second equation, \(\log_3(2x - 9) = 3\), we substitute \(x = -6\) into the argument of the logarithm:
\[2(-6) - 9 = -12 - 9 = -21.\]
Since the argument of a logarithm must be positive, \(-21\) is not a valid input, making this equation undefined for \(x = -6\).

  1. Evaluating the third equation: \(\log_3 216 = x\)

For the third equation, \(\log_3 216 = x\), substituting \(x = -6\) leads to the expression \(\log_3 216 = -6\). Converting this to exponential form, this would imply \(3^{-6} = 216\). However, \(3^{-6} = \frac{1}{3^6} = \frac{1}{729}\), which is clearly not \(216\). Thus, this equation is incorrect.

  1. Evaluating the fourth equation: \(\log_3(-2x - 3) = 2\) and Conclusion

For the fourth equation, \(\log_3(-2x - 3) = 2\), we substitute \(x = -6\) into the argument:
\[-2(-6) - 3 = 12 - 3 = 9.\]
Since \(9 > 0\), the argument is valid. Now we check if the equation holds:
\[\log_3 9 = 2.\]
Recalling that \(3^2 = 9\), this statement is correct. Thus, \(x = -6\) is indeed the solution to this equation.

Examples

Logarithmic equations are used in various fields such as chemistry for calculating pH levels, in finance for determining the time required for an investment to grow, and in seismology for measuring the magnitude of earthquakes on the Richter scale. Understanding the domain and validity of these equations ensures that we avoid impossible results in real-world measurements and predictions. For example, when calculating the pH of a solution using \(\text{pH} = -\log_{10}[H^+]\), knowing the range of possible concentrations is vital for accuracy.

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