Given `x = log_(10)12, y = log_(4)2 xx log_(10)9 and z = log_(10) 0.4`, find :
(i) x - y - z
(ii) `13^(x - y - z)`

Answer

To solve the problem step by step, we need to evaluate the expressions for \( x \), \( y \), and \( z \) and then compute \( x - y - z \) and \( 13^{(x - y - z)} \).

### Step 1: Calculate \( x \)

Given:
\[ x = \log_{10} 12 \]
We can express 12 as:
\[ 12 = 3 \times 4 \]
Using the logarithm property \( \log_a (b \times c) = \log_a b + \log_a c \), we have:
\[ x = \log_{10} 3 + \log_{10} 4 \]
Next, we can express 4 as \( 2^2 \):
\[ \log_{10} 4 = \log_{10} (2^2) = 2 \log_{10} 2 \]
Thus, we can rewrite \( x \):
\[ x = \log_{10} 3 + 2 \log_{10} 2 \]

### Step 2: Calculate \( y \)

Given:
\[ y = \log_{4} 2 \times \log_{10} 9 \]
We can express \( \log_{4} 2 \) using the change of base formula:
\[ \log_{4} 2 = \frac{\log_{10} 2}{\log_{10} 4} \]
And since \( \log_{10} 4 = 2 \log_{10} 2 \):
\[ \log_{4} 2 = \frac{\log_{10} 2}{2 \log_{10} 2} = \frac{1}{2} \]
Now, we can express \( \log_{10} 9 \) as:
\[ \log_{10} 9 = \log_{10} (3^2) = 2 \log_{10} 3 \]
Thus, substituting back into \( y \):
\[ y = \frac{1}{2} \times 2 \log_{10} 3 = \log_{10} 3 \]

### Step 3: Calculate \( z \)

Given:
\[ z = \log_{10} 0.4 \]
We can express 0.4 as:
\[ 0.4 = \frac{4}{10} = \frac{2^2}{10} = \log_{10} 4 - \log_{10} 10 \]
Since \( \log_{10} 10 = 1 \):
\[ z = \log_{10} 4 - 1 \]
And we know \( \log_{10} 4 = 2 \log_{10} 2 \):
\[ z = 2 \log_{10} 2 - 1 \]

### Step 4: Calculate \( x - y - z \)

Now we can compute:
\[ x - y - z = (\log_{10} 3 + 2 \log_{10} 2) - \log_{10} 3 - (2 \log_{10} 2 - 1) \]
Distributing the negative sign:
\[ = \log_{10} 3 + 2 \log_{10} 2 - \log_{10} 3 - 2 \log_{10} 2 + 1 \]
The \( \log_{10} 3 \) and \( 2 \log_{10} 2 \) terms cancel out:
\[ = 1 \]

### Step 5: Calculate \( 13^{(x - y - z)} \)

Now we need to compute:
\[ 13^{(x - y - z)} = 13^1 = 13 \]

### Final Answers
(i) \( x - y - z = 1 \)
(ii) \( 13^{(x - y - z)} = 13 \)