If `log_ax=alpha,log_bx=beta,log_c x=gamma` and `log_d x=delta ,xne1` and a,b,c,`dne1`,then `log_(abcd)` x equals

Answer

To solve the problem, we need to find the value of \( \log_{(abcd)} x \) given the logarithmic relationships:

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

### Step-by-Step Solution:

**Step 1: Rewrite the logarithm using the change of base formula.**

We know that:
\[ \log_{(abcd)} x = \frac{\log x}{\log (abcd)} \]

**Hint:** Remember that the change of base formula allows us to express the logarithm in terms of a different base.

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**Step 2: Simplify \( \log (abcd) \).**

Using the property of logarithms that states \( \log (A \cdot B) = \log A + \log B \), we can write:
\[ \log (abcd) = \log a + \log b + \log c + \log d \]

**Hint:** This property is very useful for combining logarithmic expressions.

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**Step 3: Express each logarithm in terms of \( x \).**

From the given information, we can express \( \log a, \log b, \log c, \) and \( \log d \) in terms of \( \alpha, \beta, \gamma, \) and \( \delta \):
\[ \log a = \frac{1}{\alpha}, \quad \log b = \frac{1}{\beta}, \quad \log c = \frac{1}{\gamma}, \quad \log d = \frac{1}{\delta} \]

**Hint:** Use the relationship \( \log_a x = \alpha \) to find \( \log a \) in terms of \( \alpha \).

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**Step 4: Substitute back into the equation for \( \log (abcd) \).**

Now substituting these values into the expression for \( \log (abcd) \):
\[ \log (abcd) = \frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta} \]

**Hint:** Make sure to keep track of the denominators when substituting values.

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**Step 5: Substitute into the expression for \( \log_{(abcd)} x \).**

Now we can substitute this back into our original equation:
\[ \log_{(abcd)} x = \frac{\log x}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}} \]

**Hint:** This step combines everything into one expression.

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**Step 6: Express \( \log x \) in terms of \( \alpha, \beta, \gamma, \delta \).**

Using the relationships again, we can express \( \log x \) as:
\[ \log x = \alpha \log a = \beta \log b = \gamma \log c = \delta \log d \]
For simplicity, we can express \( \log x \) in terms of any of these variables, but we will use \( \log x = \alpha \log a \).

**Hint:** Choose one of the logarithmic expressions to express \( \log x \).

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**Final Step: Combine everything to find the final expression.**

Thus, we have:
\[ \log_{(abcd)} x = \frac{\alpha \log a}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}} \]

This can be simplified further, but the key result is:
\[ \log_{(abcd)} x = \frac{1}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}} \]

### Final Result:
\[ \log_{(abcd)} x = \frac{1}{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma} + \frac{1}{\delta}} \]