IF `x = log_(2a)a, y = log_(3a) 2a , z = log_(4a)3a`, then the value of `xyz + 1` is

Options

A. 2yz
B. 2zx
C. xyz
D. 1

Answer

To solve the problem, we need to find the value of \( xyz + 1 \) where:

\[ x = \log_{2a} a, \quad y = \log_{3a} 2a, \quad z = \log_{4a} 3a \]

### Step 1: Rewrite the logarithms using the change of base formula

Using the change of base formula, we can express each logarithm in terms of natural logarithms (ln):

\[ x = \frac{\ln a}{\ln(2a)} = \frac{\ln a}{\ln 2 + \ln a} \]

\[ y = \frac{\ln(2a)}{\ln(3a)} = \frac{\ln 2 + \ln a}{\ln 3 + \ln a} \]

\[ z = \frac{\ln(3a)}{\ln(4a)} = \frac{\ln 3 + \ln a}{\ln 4 + \ln a} \]

### Step 2: Substitute \( x, y, z \) into \( xyz + 1 \)

Now we can substitute these expressions into \( xyz \):

\[ xyz = \left(\frac{\ln a}{\ln 2 + \ln a}\right) \left(\frac{\ln 2 + \ln a}{\ln 3 + \ln a}\right) \left(\frac{\ln 3 + \ln a}{\ln 4 + \ln a}\right) \]

Notice that the \( \ln 2 + \ln a \) in the numerator of \( y \) cancels with the denominator of \( x \):

\[ xyz = \frac{\ln a}{\ln 3 + \ln a} \cdot \frac{\ln 3 + \ln a}{\ln 4 + \ln a} \]

Now, the \( \ln 3 + \ln a \) in the numerator of \( z \) cancels with the denominator of the previous fraction:

\[ xyz = \frac{\ln a}{\ln 4 + \ln a} \]

### Step 3: Add 1 to \( xyz \)

Now we need to compute \( xyz + 1 \):

\[ xyz + 1 = \frac{\ln a}{\ln 4 + \ln a} + 1 \]

To add these fractions, we need a common denominator:

\[ = \frac{\ln a + (\ln 4 + \ln a)}{\ln 4 + \ln a} = \frac{\ln a + \ln 4 + \ln a}{\ln 4 + \ln a} = \frac{2\ln a + \ln 4}{\ln 4 + \ln a} \]

### Step 4: Simplify the expression

Notice that \( \ln 4 = \ln(2^2) = 2\ln 2 \):

\[ = \frac{2\ln a + 2\ln 2}{\ln 4 + \ln a} = \frac{2(\ln a + \ln 2)}{\ln 4 + \ln a} \]

### Final Result

Thus, the expression simplifies to:

\[ xyz + 1 = \frac{2\ln(2a)}{\ln(4a)} \]

This concludes the solution.