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\\)\
done

Explanation:

Step1: Check $\log_{3}(2x + 15)=2$

Using the definition of logarithm: if $\log_{a}b = c$, then $b=a^{c}$. So for $\log_{3}(2x + 15)=2$, we have $2x + 15 = 3^{2}=9$. Solving for $x$: $2x=9 - 15=-6$, so $x=-3$.

Step2: Check $\log_{5}(8x + 9)=2$

Using the logarithm definition: $8x + 9 = 5^{2}=25$. Solving for $x$: $8x=25 - 9 = 16$, so $x = 2
eq-3$.

Step3: Check $\log_{4}(-20x + 4)=3$

Using the logarithm definition: $-20x + 4=4^{3}=64$. Solving for $x$: $-20x=64 - 4 = 60$, so $x=-3$.

Step4: Check $\log_{x}81 = 4$

Using the logarithm definition: $x^{4}=81$. Solving for $x$: $x=\pm3$, but the base of a logarithm must be positive and not equal to 1, so $x = 3$ (since $x=-3$ is invalid as a base), so $x
eq - 3$.

Answer:

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