Given `log_(a) x=alpha, log _(b) x=beta, log_(c)x =gamma & log_(d) x=delta(x ne 1), a,b,c,d in R^(+)-{1}" then "log_(abcd)x` then the value equal to:

Options

A. `(1)/(alphabeta gamma delta)`
B. `(1)/(alpha)+(1)/(beta)+(1)/(gamma)+(1)/(delta)`
C. `(1)/((1)/(alpha)+(1)/(beta)+(1)/(gamma)+(1)/(delta))`
D. None of these

Answer

To solve the problem, we start with the given logarithmic equations:

1. \( \log_a x = \alpha \)
2. \( \log_b x = \beta \)
3. \( \log_c x = \gamma \)
4. \( \log_d x = \delta \)

We need to find the value of \( \log_{abcd} x \).

### Step 1: Express the logarithms in terms of their reciprocals
Using the property of logarithms that states \( \log_a b = \frac{1}{\log_b a} \), we can express the logarithms in terms of their reciprocals:

\[ \log_a x = \alpha \implies \log_x a = \frac{1}{\alpha} \]
\[ \log_b x = \beta \implies \log_x b = \frac{1}{\beta} \]
\[ \log_c x = \gamma \implies \log_x c = \frac{1}{\gamma} \]
\[ \log_d x = \delta \implies \log_x d = \frac{1}{\delta} \]

### Step 2: Add the logarithmic values
Now, we can add these four logarithmic values:

\[ \log_x a + \log_x b + \log_x c + \log_x d = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} \]

### Step 3: Use the property of logarithms
Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \), we can combine the left side:

\[ \log_x (abcd) = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} \]

### Step 4: Change the base
Now, we want to express \( \log_{abcd} x \). We can use the change of base formula:

\[ \log_{abcd} x = \frac{1}{\log_x (abcd)} \]

Substituting the expression we found in Step 3:

\[ \log_{abcd} x = \frac{1}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}} \]

### Final Result
Thus, the value of \( \log_{abcd} x \) is:

\[ \log_{abcd} x = \frac{1}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}} \]