`log_(2)[log_(4)(log_(10)16^(4)+log_(10)25^(8))]` simplifies to :

Options

A. an irrational
B. an odd prime
C. a composite
D. unity

Answer

To simplify the expression \( \log_{2}\left[\log_{4}\left(\log_{10}(16^{4}) + \log_{10}(25^{8})\right)\right] \), we will follow these steps:

### Step 1: Simplify the logarithmic terms inside
We start with the expression inside the logarithm:
\[ \log_{10}(16^{4}) + \log_{10}(25^{8}) \]
Using the property of logarithms \( \log_{a}(b^{c}) = c \cdot \log_{a}(b) \):
\[ \log_{10}(16^{4}) = 4 \cdot \log_{10}(16) \quad \text{and} \quad \log_{10}(25^{8}) = 8 \cdot \log_{10}(25) \]
Now, we know that \( 16 = 2^{4} \) and \( 25 = 5^{2} \):
\[ \log_{10}(16) = \log_{10}(2^{4}) = 4 \cdot \log_{10}(2) \quad \text{and} \quad \log_{10}(25) = \log_{10}(5^{2}) = 2 \cdot \log_{10}(5) \]
Substituting these back, we get:
\[ \log_{10}(16^{4}) = 4 \cdot (4 \cdot \log_{10}(2)) = 16 \cdot \log_{10}(2) \]
\[ \log_{10}(25^{8}) = 8 \cdot (2 \cdot \log_{10}(5)) = 16 \cdot \log_{10}(5) \]
Thus, we can combine these:
\[ \log_{10}(16^{4}) + \log_{10}(25^{8}) = 16 \cdot \log_{10}(2) + 16 \cdot \log_{10}(5) = 16(\log_{10}(2) + \log_{10}(5)) \]
Using the property \( \log_{a}(b) + \log_{a}(c) = \log_{a}(bc) \):
\[ \log_{10}(2) + \log_{10}(5) = \log_{10}(10) = 1 \]
So, we have:
\[ \log_{10}(16^{4}) + \log_{10}(25^{8}) = 16 \cdot 1 = 16 \]

### Step 2: Substitute back into the logarithmic expression
Now we substitute this back into the original expression:
\[ \log_{2}\left[\log_{4}(16)\right] \]

### Step 3: Simplify \( \log_{4}(16) \)
Next, we simplify \( \log_{4}(16) \):
We know that \( 16 = 4^{2} \), so:
\[ \log_{4}(16) = \log_{4}(4^{2}) = 2 \]

### Step 4: Substitute and simplify \( \log_{2}(2) \)
Now we substitute this back into the expression:
\[ \log_{2}(2) \]
Using the property \( \log_{a}(a) = 1 \):
\[ \log_{2}(2) = 1 \]

### Final Result
Thus, the entire expression simplifies to:
\[ \log_{2}\left[\log_{4}\left(\log_{10}(16^{4}) + \log_{10}(25^{8})\right)\right] = 1 \]

### Conclusion
The final answer is \( 1 \), which corresponds to the option "unity".

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