Question
Solve the equation log_3 (x^2 +9x +27) =2
Answers
GPT-4o mini
To solve the equation \( \log_3 (x^2 + 9x + 27) = 2 \), we first convert the logarithmic equation into an exponential form. The equation can be rewritten as:
\[
x^2 + 9x + 27 = 3^2
\]
Calculating \( 3^2 \):
\[
x^2 + 9x + 27 = 9
\]
Next, we rearrange the equation by moving all terms to one side:
\[
x^2 + 9x + 27 - 9 = 0
\]
This simplifies to:
\[
x^2 + 9x + 18 = 0
\]
Now, we can factor the quadratic:
To factor the quadratic \( x^2 + 9x + 18 \), we need two numbers that multiply to 18 and add up to 9. The numbers 3 and 6 work:
\[
(x + 3)(x + 6) = 0
\]
Setting each factor to zero gives us the solutions:
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 6 = 0 \quad \Rightarrow \quad x = -6
\]
Thus, the solutions to the equation are:
\[
x = -3 \quad \text{and} \quad x = -6
\]
Finally, we need to check if these solutions are valid in the original logarithmic expression, since the argument of the logarithm must be positive.
**Checking \( x = -3 \)**:
\[
x^2 + 9x + 27 = (-3)^2 + 9(-3) + 27 = 9 - 27 + 27 = 9 \quad (\text{valid since } 9 > 0)
\]
**Checking \( x = -6 \)**:
\[
x^2 + 9x + 27 = (-6)^2 + 9(-6) + 27 = 36 - 54 + 27 = 9 \quad (\text{valid since } 9 > 0)
\]
Since both solutions yield a positive argument for the logarithm, they are both valid.
Thus, the final solutions are:
\[
\boxed{-3} \quad \text{and} \quad \boxed{-6}
\]
\[
x^2 + 9x + 27 = 3^2
\]
Calculating \( 3^2 \):
\[
x^2 + 9x + 27 = 9
\]
Next, we rearrange the equation by moving all terms to one side:
\[
x^2 + 9x + 27 - 9 = 0
\]
This simplifies to:
\[
x^2 + 9x + 18 = 0
\]
Now, we can factor the quadratic:
To factor the quadratic \( x^2 + 9x + 18 \), we need two numbers that multiply to 18 and add up to 9. The numbers 3 and 6 work:
\[
(x + 3)(x + 6) = 0
\]
Setting each factor to zero gives us the solutions:
\[
x + 3 = 0 \quad \Rightarrow \quad x = -3
\]
\[
x + 6 = 0 \quad \Rightarrow \quad x = -6
\]
Thus, the solutions to the equation are:
\[
x = -3 \quad \text{and} \quad x = -6
\]
Finally, we need to check if these solutions are valid in the original logarithmic expression, since the argument of the logarithm must be positive.
**Checking \( x = -3 \)**:
\[
x^2 + 9x + 27 = (-3)^2 + 9(-3) + 27 = 9 - 27 + 27 = 9 \quad (\text{valid since } 9 > 0)
\]
**Checking \( x = -6 \)**:
\[
x^2 + 9x + 27 = (-6)^2 + 9(-6) + 27 = 36 - 54 + 27 = 9 \quad (\text{valid since } 9 > 0)
\]
Since both solutions yield a positive argument for the logarithm, they are both valid.
Thus, the final solutions are:
\[
\boxed{-3} \quad \text{and} \quad \boxed{-6}
\]