To solve this equation:

$\frac{\log_3(3x + 6)}{\log_3(81)} = 1,$

we apply the properties of logarithms step by step:

Step 1: Simplify $\log_3(81)$

We know that $81 = 3^4$, so:

$\log_3(81) = 4.$

The equation now becomes:

$\frac{\log_3(3x + 6)}{4} = 1.$

Step 2: Eliminate the denominator

Multiply through by 4 to eliminate the denominator:

$\log_3(3x + 6) = 4.$

Step 3: Rewrite in exponential form

Using the property $\log_b(a) = c \implies b^c = a$, rewrite the logarithmic equation:

$3^4 = 3x + 6.$

Simplify $3^4$:

$81 = 3x + 6.$

Step 4: Solve for $x$

Subtract 6 from both sides:

$75 = 3x.$

Divide by 3:

$x = 25.$

Final Answer:

$x = 25$

Let me know if you'd like further clarification!

Related Questions:

1. What are the general properties of logarithms used here?
2. How do you rewrite logarithmic equations in exponential form?
3. Why is $\log_3(81) = 4$? Explain step-by-step.
4. Can the solution $x = 25$ be verified? If so, how?
5. What would the equation look like if $\log_3(3x + 6) = 5$?

Tip:

Always check your solutions by substituting back into the original equation to verify correctness.