which equation has x = -6 as the solution?\nlog_x 36 = 2\nlog_3(2x - 9) = 3\nlog_3 216 = x\nlog_3(-2x - 3) = 2

Explanation:

Step1: Recall the definition of logarithms

If $\log_{a}b = c$, then $a^{c}=b$.

Step2: Check $\log_{x}36 = 2$

By the definition, $x^{2}=36$, so $x=\pm6$. But the base of a logarithm $x>0,x\neq1$, so $x = 6$ is valid and $x=-6$ is not.

Step3: Check $\log_{3}(2x - 9)=3$

Using the definition, $3^{3}=2x - 9$. Then $27=2x - 9$. Add 9 to both sides: $2x=27 + 9=36$, so $x = 18$.

Step4: Check $\log_{3}216=x$

By the definition, $3^{x}=216$. Since $3^{3}=27,3^{4}=81,3^{5}=243$, this equation does not have $x=-6$ as a solution.

Step5: Check $\log_{3}(-2x - 3)=2$

Using the definition, $3^{2}=-2x - 3$. So $9=-2x - 3$. Add 3 to both sides: $12=-2x$. Divide by - 2, we get $x=-6$.

Answer:

$\log_{3}(-2x - 3)=2$