If p = log 20 and q = log 25, find the value of x, if 2 log(x + 1) = 2p - q.

Answer

To solve the equation \( 2 \log(x + 1) = 2p - q \) given \( p = \log 20 \) and \( q = \log 25 \), we will follow these steps:

### Step 1: Substitute the values of \( p \) and \( q \)
We start by substituting \( p \) and \( q \) into the equation:

\[ 2 \log(x + 1) = 2 \log 20 - \log 25 \]

### Step 2: Use the properties of logarithms
We can simplify the right-hand side using the properties of logarithms. Recall that \( a \log b = \log(b^a) \) and \( \log a - \log b = \log\left(\frac{a}{b}\right) \):

\[ 2 \log 20 = \log(20^2) = \log 400 \]
Thus, we can rewrite the equation as:

\[ 2 \log(x + 1) = \log 400 - \log 25 \]

### Step 3: Combine the logarithms
Now, we can combine the logarithms on the right side:

\[ \log 400 - \log 25 = \log\left(\frac{400}{25}\right) \]

Calculating \( \frac{400}{25} \):

\[ \frac{400}{25} = 16 \]

So, we have:

\[ 2 \log(x + 1) = \log 16 \]

### Step 4: Simplify the left-hand side
Using the property of logarithms again, we can rewrite the left-hand side:

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

Thus, we can equate both sides:

\[ \log((x + 1)^2) = \log 16 \]

### Step 5: Remove the logarithm
Since the logarithm is equal on both sides, we can remove the logarithm (assuming the arguments are positive):

\[ (x + 1)^2 = 16 \]

### Step 6: Solve for \( x + 1 \)
Taking the square root of both sides gives us:

\[ x + 1 = 4 \quad \text{or} \quad x + 1 = -4 \]

Since \( x + 1 \) must be positive (as it is inside a logarithm), we take:

\[ x + 1 = 4 \]

### Step 7: Solve for \( x \)
Now, we can solve for \( x \):

\[ x = 4 - 1 = 3 \]

### Final Answer
Thus, the value of \( x \) is:

\[ \boxed{3} \]