Solve: `log_axlog_a(xyz)=48`; `log_aylog_a(xyz)=12`; `log_azlog_a(xyz)=84`

Explanation

To solve the equations given in the problem, we will follow these steps:

### Step 1: Write down the given equations
We have the following equations:
1. \( \log_a x \cdot \log_a (xyz) = 48 \)
2. \( \log_a y \cdot \log_a (xyz) = 12 \)
3. \( \log_a z \cdot \log_a (xyz) = 84 \)

### Step 2: Introduce a new variable
Let \( k = \log_a (xyz) \). Then we can rewrite the equations as:
1. \( \log_a x \cdot k = 48 \) (Equation 1)
2. \( \log_a y \cdot k = 12 \) (Equation 2)
3. \( \log_a z \cdot k = 84 \) (Equation 3)

### Step 3: Solve for \( \log_a x \), \( \log_a y \), and \( \log_a z \)
From Equation 1:
\[ \log_a x = \frac{48}{k} \]

From Equation 2:
\[ \log_a y = \frac{12}{k} \]

From Equation 3:
\[ \log_a z = \frac{84}{k} \]

### Step 4: Add the logarithmic equations
Now, we can add the three logarithmic values:
\[ \log_a x + \log_a y + \log_a z = \frac{48}{k} + \frac{12}{k} + \frac{84}{k} \]
This simplifies to:
\[ \log_a (xyz) = \frac{48 + 12 + 84}{k} = \frac{144}{k} \]

### Step 5: Set up the equation
Since we know that \( \log_a (xyz) = k \), we can equate:
\[ k = \frac{144}{k} \]

### Step 6: Solve for \( k \)
Multiplying both sides by \( k \) gives:
\[ k^2 = 144 \]
Taking the square root of both sides, we find:
\[ k = 12 \quad \text{(since logarithms are positive)} \]

### Step 7: Substitute back to find \( \log_a x \), \( \log_a y \), and \( \log_a z \)
Now substitute \( k = 12 \) back into the equations:
1. \( \log_a x = \frac{48}{12} = 4 \)
2. \( \log_a y = \frac{12}{12} = 1 \)
3. \( \log_a z = \frac{84}{12} = 7 \)

### Step 8: Convert logarithmic values to exponential form
Now we can convert these logarithmic values back to their exponential forms:
1. \( x = a^4 \)
2. \( y = a^1 = a \)
3. \( z = a^7 \)

### Step 9: Find \( x - 5z \)
Now we need to find \( x - 5z \):
\[ x - 5z = a^4 - 5(a^7) = a^4 - 5a^7 \]

### Final Result
Thus, the final expression for \( x - 5z \) is:
\[ x - 5z = a^4 - 5a^7 \]