if `log_2(log_3(log_4x))=0` and `log_3(log_4(log_2y))=0`and `log_3(log_2(log_3z))=0` then find the sum of `x, y` and `z` is

Options

A. 89
B. 58
C. 105
D. 49

Answer

To solve the equations given in the problem, we will break it down step by step.

### Step 1: Solve for x
We start with the equation:
\[ \log_2(\log_3(\log_4 x)) = 0 \]
Since \(\log_a b = 0\) implies \(b = 1\), we can rewrite the equation as:
\[ \log_3(\log_4 x) = 1 \]
Now, we know that \(\log_3 b = 1\) implies \(b = 3\). Therefore:
\[ \log_4 x = 3 \]
Using the property of logarithms, we can convert this to exponential form:
\[ x = 4^3 = 64 \]

### Step 2: Solve for y
Next, we consider the second equation:
\[ \log_3(\log_4(\log_2 y)) = 0 \]
Again, using the same reasoning as before:
\[ \log_4(\log_2 y) = 1 \]
This implies:
\[ \log_2 y = 4 \]
Converting this to exponential form gives us:
\[ y = 2^4 = 16 \]

### Step 3: Solve for z
Now, we move on to the third equation:
\[ \log_3(\log_2(\log_3 z)) = 0 \]
Following the same steps:
\[ \log_2(\log_3 z) = 1 \]
This implies:
\[ \log_3 z = 2 \]
Converting this to exponential form gives us:
\[ z = 3^2 = 9 \]

### Step 4: Calculate the sum of x, y, and z
Now that we have found \(x\), \(y\), and \(z\):
\[ x = 64, \quad y = 16, \quad z = 9 \]
We can find the sum:
\[ x + y + z = 64 + 16 + 9 = 89 \]

### Final Answer
Thus, the sum of \(x\), \(y\), and \(z\) is:
\[ \boxed{89} \]