If `a^x=b^y=c^z=d^w` then `log_a(bcd)=`

Options

A. `z((1)/(x)+(1)/(y)+(1)/(w))`
B. `y((1)/(x)+(1)/(z)+(1)/(w))`
C. `x((1)/(y)+(1)/(z)+(1)/(w))`
D. `(xyz)/(w)`

Answer

To solve the problem \( \log_a(bcd) \) given that \( a^x = b^y = c^z = d^w \), we can follow these steps:

### Step 1: Set a common variable
Let \( k = a^x = b^y = c^z = d^w \). This means we can express \( a, b, c, \) and \( d \) in terms of \( k \):
- From \( a^x = k \), we have \( a = k^{1/x} \).
- From \( b^y = k \), we have \( b = k^{1/y} \).
- From \( c^z = k \), we have \( c = k^{1/z} \).
- From \( d^w = k \), we have \( d = k^{1/w} \).

### Step 2: Substitute \( b, c, d \) into \( \log_a(bcd) \)
Now we can substitute \( b, c, \) and \( d \) into the logarithmic expression:
\[ bcd = b \cdot c \cdot d = k^{1/y} \cdot k^{1/z} \cdot k^{1/w} \]
Using the property of exponents, we can combine these:
\[ bcd = k^{(1/y) + (1/z) + (1/w)} \]

### Step 3: Write the logarithm
Now we need to evaluate:
\[ \log_a(bcd) = \log_a(k^{(1/y) + (1/z) + (1/w)}) \]

### Step 4: Use the logarithmic identity
Using the property of logarithms \( \log_b(m^n) = n \cdot \log_b(m) \):
\[ \log_a(bcd) = \left( \frac{1}{y} + \frac{1}{z} + \frac{1}{w} \right) \cdot \log_a(k) \]

### Step 5: Change of base for \( \log_a(k) \)
We know that \( a = k^{1/x} \), so:
\[ \log_a(k) = \log_{k^{1/x}}(k) = \frac{1}{(1/x)} \cdot \log_k(k) = x \]
since \( \log_k(k) = 1 \).

### Step 6: Substitute back into the expression
Now substituting back:
\[ \log_a(bcd) = \left( \frac{1}{y} + \frac{1}{z} + \frac{1}{w} \right) \cdot x \]

### Final Answer
Thus, the final answer is:
\[ \log_a(bcd) = x \left( \frac{1}{y} + \frac{1}{z} + \frac{1}{w} \right) \]