We are given the equation:

\[
\log(x+5) + \log(x-5) = 4\log2 + 2\log3
\]

### Step 1: Use the properties of logarithms

1. **Sum of logarithms**: Apply the property \(\log a + \log b = \log(ab)\) to the left-hand side:
   \[
   \log((x+5)(x-5)) = 4\log2 + 2\log3
   \]

2. **Power of logarithms**: Use the property \(k\log a = \log(a^k)\) on the right-hand side:
   \[
   4\log2 + 2\log3 = \log(2^4) + \log(3^2) = \log(16) + \log(9)
   \]

3. **Sum of logarithms** (again):
   \[
   \log(16) + \log(9) = \log(16 \times 9) = \log(144)
   \]

So, the equation becomes:
\[
\log((x+5)(x-5)) = \log(144)
\]

### Step 2: Remove the logarithms

Since both sides of the equation have logarithms with the same base, we can equate the arguments:
\[
(x+5)(x-5) = 144
\]

### Step 3: Expand and solve

Use the difference of squares formula on the left-hand side:
\[
x^2 - 25 = 144
\]

Add 25 to both sides:
\[
x^2 = 169
\]

Take the square root of both sides:
\[
x = \pm 13
\]

### Step 4: Check for valid solutions

Since \( \log(x+5) \) and \( \log(x-5) \) must both be defined (i.e., their arguments must be positive):
- If \( x = 13 \), then \( x+5 = 18 \) and \( x-5 = 8 \), which are both positive.
- If \( x = -13 \), then \( x+5 = -8 \), which is not valid because we cannot take the logarithm of a negative number.

Thus, the only valid solution is:
\[
x = 13
\]

Let me know if you'd like any further details or clarification!

Here are some related questions you might be interested in:

1. How can the properties of logarithms simplify other equations?
2. What is the difference of squares formula, and how is it applied?
3. Can you solve logarithmic equations where the bases are different?
4. How would the equation change if one side had a natural logarithm (\(\ln\))?
5. What happens if \(x\) takes negative values in logarithmic functions?

**Tip**: When solving logarithmic equations, always check if the arguments of the logarithms are positive to ensure the solution is valid!

Find x if, log(x+5) + log(x-5) = 4log2 + 2log3

Explore how to solve the logarithmic equation log(x+5) + log(x-5) = 4log2 + 2log3 by applying properties of logarithms and quadratic equations. This solution covers key algebraic concepts, including the sum and power of logarithms, and is suitable for students in Grades 10-12.

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