There exists a positive number k such that ` log_2x+ log_4x+ log_8x= log_kx`, for all positive real no x. If k=`a^(1/b)` where (a,b) `epsilon` N, the smallest possible value of (a+b)= (C) 12 A) 75 (B) 65 (D) 63_

Options

A. 12
B. 63
C. 65
D. 75

Answer

To solve the equation \( \log_2 x + \log_4 x + \log_8 x = \log_k x \) for a positive number \( k \), we will follow these steps:

### Step 1: Rewrite the logarithms in terms of base 2
We can express \( \log_4 x \) and \( \log_8 x \) in terms of \( \log_2 x \):
\[ \log_4 x = \log_{2^2} x = \frac{1}{2} \log_2 x \]
\[ \log_8 x = \log_{2^3} x = \frac{1}{3} \log_2 x \]

### Step 2: Substitute these values into the equation
Substituting the values we found into the original equation:
\[ \log_2 x + \frac{1}{2} \log_2 x + \frac{1}{3} \log_2 x = \log_k x \]

### Step 3: Combine the logarithmic terms
Now, we can combine the left-hand side:
\[ \left(1 + \frac{1}{2} + \frac{1}{3}\right) \log_2 x = \log_k x \]

### Step 4: Find a common denominator and simplify
The common denominator of 1, 2, and 3 is 6. Thus, we can rewrite the sum:
\[ 1 = \frac{6}{6}, \quad \frac{1}{2} = \frac{3}{6}, \quad \frac{1}{3} = \frac{2}{6} \]
Adding these together:
\[ \frac{6 + 3 + 2}{6} = \frac{11}{6} \]
So we have:
\[ \frac{11}{6} \log_2 x = \log_k x \]

### Step 5: Use the property of logarithms to equate
Using the property of logarithms, we can express \( \log_k x \) as:
\[ \log_k x = \frac{\log_2 x}{\log_2 k} \]
Thus, we can equate:
\[ \frac{11}{6} \log_2 x = \frac{\log_2 x}{\log_2 k} \]

### Step 6: Cancel \( \log_2 x \) (assuming \( x > 0 \))
Since \( x \) is a positive real number, we can safely divide both sides by \( \log_2 x \):
\[ \frac{11}{6} = \frac{1}{\log_2 k} \]

### Step 7: Solve for \( \log_2 k \)
Taking the reciprocal gives:
\[ \log_2 k = \frac{6}{11} \]

### Step 8: Express \( k \) in terms of powers of 2
Now, we can express \( k \):
\[ k = 2^{\frac{6}{11}} = 2^{6/11} \]

### Step 9: Identify \( a \) and \( b \)
We can express \( k \) in the form \( a^{1/b} \):
\[ k = (2^6)^{1/11} = 64^{1/11} \]
Here, \( a = 64 \) and \( b = 11 \).

### Step 10: Calculate \( a + b \)
Now we find \( a + b \):
\[ a + b = 64 + 11 = 75 \]

### Final Answer
Thus, the smallest possible value of \( a + b \) is \( \boxed{75} \).