`(log (x^(3) + 3x^(2) + 3x + 1))/(log (x^(2) + 2x + 1))` is equal to

Options

A. `(1)/(2)`
B. 1
C. `(3)/(2)`
D. `(2)/(3)`

Answer

To solve the expression \(\frac{\log(x^3 + 3x^2 + 3x + 1)}{\log(x^2 + 2x + 1)}\), we will simplify both the numerator and the denominator.

### Step 1: Simplify the numerator
The numerator is \(\log(x^3 + 3x^2 + 3x + 1)\). We can recognize that the expression \(x^3 + 3x^2 + 3x + 1\) can be factored as \((x + 1)^3\).

\[ x^3 + 3x^2 + 3x + 1 = (x + 1)^3 \]

### Step 2: Apply the logarithmic property
Using the property of logarithms that states \(\log(a^b) = b \cdot \log(a)\), we can rewrite the numerator:

\[ \log((x + 1)^3) = 3 \cdot \log(x + 1) \]

### Step 3: Simplify the denominator
Now, let's simplify the denominator, which is \(\log(x^2 + 2x + 1)\). We can recognize that \(x^2 + 2x + 1\) can be factored as \((x + 1)^2\).

\[ x^2 + 2x + 1 = (x + 1)^2 \]

### Step 4: Apply the logarithmic property to the denominator
Using the same logarithmic property, we rewrite the denominator:

\[ \log((x + 1)^2) = 2 \cdot \log(x + 1) \]

### Step 5: Substitute back into the original expression
Now we can substitute the simplified forms of the numerator and denominator back into the original expression:

\[ \frac{\log(x^3 + 3x^2 + 3x + 1)}{\log(x^2 + 2x + 1)} = \frac{3 \cdot \log(x + 1)}{2 \cdot \log(x + 1)} \]

### Step 6: Cancel out the common terms
Since \(\log(x + 1)\) is common in both the numerator and the denominator, we can cancel it out (assuming \(x + 1 \neq 1\)):

\[ = \frac{3}{2} \]

### Final Answer
Thus, the value of the expression is:

\[ \frac{3}{2} \]