Question

which equation(s) have $x = -3$ as the solution? choose two correct answers. $\log_{4}(-20x + 4) = 3$ $\log_{x}81 = 4$ $\log_{5}(8x + 9) = 2$ $\log_{3}(2x + 15) = 2$

Explanation:

Step1: Test $x=-3$ in first equation

Substitute $x=-3$ into $\log_4(-20x + 4)$:
$\log_4(-20\times(-3)+4)=\log_4(60+4)=\log_4(64)$
Since $4^3=64$, $\log_4(64)=3$, which matches the right-hand side.

Step2: Test $x=-3$ in second equation

Substitute $x=-3$ into $\log_x 81$:
Logarithm base cannot be negative, so $x=-3$ is invalid here.

Step3: Test $x=-3$ in third equation

Substitute $x=-3$ into $\log_5(8x + 9)$:
$\log_5(8\times(-3)+9)=\log_5(-24+9)=\log_5(-15)$
Logarithm argument cannot be negative, so $x=-3$ is invalid here.

Step4: Test $x=-3$ in fourth equation

Substitute $x=-3$ into $\log_3(2x + 15)$:
$\log_3(2\times(-3)+15)=\log_3(-6+15)=\log_3(9)$
Since $3^2=9$, $\log_3(9)=2$, which matches the right-hand side.

Answer:

$\log_4(-20x + 4) = 3$, $\log_3(2x + 15) = 2$