`log_(a)b = log_(b)c = log_(c)a, ` then a, b and c are such that

Options

A. `a = b = c`
B. `a = b ne c`
C. `a = c ne b`
D. `b = c ne a `

Answer

To solve the problem given by the equation \( \log_a b = \log_b c = \log_c a \), we will denote the common value of these logarithms as \( k \). This means we can write:

1. \( \log_a b = k \)
2. \( \log_b c = k \)
3. \( \log_c a = k \)

### Step 1: Express \( b \), \( c \), and \( a \) in terms of \( k \)

Using the property of logarithms, we can convert these logarithmic equations into exponential form:

1. From \( \log_a b = k \), we have:
\[ b = a^k \]

2. From \( \log_b c = k \), we have:
\[ c = b^k \]
Substituting \( b = a^k \) into this equation gives:
\[ c = (a^k)^k = a^{k^2} \]

3. From \( \log_c a = k \), we have:
\[ a = c^k \]
Substituting \( c = a^{k^2} \) into this equation gives:
\[ a = (a^{k^2})^k = a^{k^3} \]

### Step 2: Set up the equation

Now we have:
\[ a = a^{k^3} \]

### Step 3: Analyze the equation

For the above equation to hold true, we can either have \( a = 0 \) (which is not valid in logarithmic terms) or the exponents must be equal. Therefore:
\[ 1 = k^3 \]

### Step 4: Solve for \( k \)

The only real solution to \( k^3 = 1 \) is:
\[ k = 1 \]

### Step 5: Substitute \( k \) back into the equations

Now substituting \( k = 1 \) back into our expressions for \( b \) and \( c \):
1. \( b = a^1 = a \)
2. \( c = a^{1^2} = a \)

### Conclusion

Thus, we find that:
\[ a = b = c \]

### Final Answer

The relationship between \( a \), \( b \), and \( c \) is:
\[ a = b = c \]