We are tasked with solving the equation:

\[
2\log_3{(x+7)} = \log_3{9} + 2
\]

### Step 1: Use logarithmic properties

We know that \(\log_b{(a^c)} = c\log_b{a}\), so we can start by simplifying the left-hand side:

\[
2\log_3{(x+7)} = \log_3{(x+7)^2}
\]

Thus, the equation becomes:

\[
\log_3{(x+7)^2} = \log_3{9} + 2
\]

### Step 2: Simplify \(\log_3{9}\)

We know that \(9 = 3^2\), so:

\[
\log_3{9} = \log_3{3^2} = 2
\]

Thus, the equation simplifies to:

\[
\log_3{(x+7)^2} = 2 + 2
\]

\[
\log_3{(x+7)^2} = 4
\]

### Step 3: Remove the logarithm by exponentiation

Since \(\log_b{y} = x\) is equivalent to \(b^x = y\), we raise both sides of the equation to powers of 3:

\[
(x+7)^2 = 3^4
\]

\[
(x+7)^2 = 81
\]

### Step 4: Solve for \(x\)

Take the square root of both sides:

\[
x+7 = \pm \sqrt{81}
\]

\[
x+7 = \pm 9
\]

Thus, we have two cases:
1. \(x+7 = 9\) ⟹ \(x = 2\)
2. \(x+7 = -9\) ⟹ \(x = -16\)

### Step 5: Check the solutions

For logarithmic equations, the argument of the logarithm must be positive. Let's check both solutions:

- For \(x = 2\), \(x + 7 = 9\), which is positive.
- For \(x = -16\), \(x + 7 = -9\), which is negative and not allowed.

Thus, the only valid solution is:

\[
\boxed{x = 2}
\]

---

Would you like more details or have any questions?

Here are 5 related questions:
1. What are the properties of logarithms used in solving logarithmic equations?
2. How do you handle logarithmic equations when the base is not explicitly given?
3. What are the conditions for the arguments of logarithmic functions to be valid?
4. How would the solution change if the equation was in base 10 instead of base 3?
5. How do you verify solutions of logarithmic equations?

**Tip**: Always check the domain of the logarithmic function to ensure the argument is positive before accepting solutions.

Learn how to solve the logarithmic equation 2log3(x+7)=log39+2 with a step-by-step explanation. This problem involves logarithmic properties, exponentiation, and checking solution validity. Suitable for high school students in Grades 10-12.

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