Question

which equation(s) have \\(x = -3\\) as the solution?

\\(\log_{3}(2x + 15) = 2\\)
\\(\log_{5}(8x + 9) = 2\\)
\\(\log_{4}(-20x + 4) = 3\\)
\\(\log_{x}81 = 4\\)

Explanation:

⚡ Using what you learned: Logarithms and Exponential Equations

Step 1: Test the first equation

Substitute \( x = -3 \) into \( \log_{3}(2x + 15) = 2 \):

$$ \log_{3}(2(-3) + 15) = \log_{3}(-6 + 15) = \log_{3}(9) $$

Since \( 3^2 = 9 \):

$$ \log_{3}(9) = 2 $$

This equation is true for \( x = -3 \).

Step 2: Test the second equation

Substitute \( x = -3 \) into \( \log_{5}(8x + 9) = 2 \):

$$ \log_{5}(8(-3) + 9) = \log_{5}(-24 + 9) = \log_{5}(-15) $$

The argument of a logarithm must be positive. Since \( -15 \le 0 \), \( \log_{5}(-15) \) is undefined.

This equation does not have \( x = -3 \) as a solution.

Step 3: Test the third equation

Substitute \( x = -3 \) into \( \log_{4}(-20x + 4) = 3 \):

$$ \log_{4}(-20(-3) + 4) = \log_{4}(60 + 4) = \log_{4}(64) $$

Since \( 4^3 = 64 \):

$$ \log_{4}(64) = 3 $$

This equation is true for \( x = -3 \).

Step 4: Test the fourth equation

Substitute \( x = -3 \) into \( \log_{x}81 = 4 \):

$$ \log_{-3}81 = 4 $$

The base of a logarithm must be positive and not equal to \( 1 \). Since the base \( -3 \) is negative, this expression is undefined.

This equation does not have \( x = -3 \) as a solution.

Answer:

• \( \log_{3}(2x + 15) = 2 \)
• \( \log_{4}(-20x + 4) = 3 \)