Nilai dari 2 log 4 + 2log 8-2log 16 adalah ....

Recall that $\log_b x = y$ means $b^y = x$.
For the first term, $^2\log 4$:
We need to find the power to which 2 must be raised to get 4.
Since $2^2 = 4$, we have $^2\log 4 = 2$.

For the second term, $^2\log 8$:
We need to find the power to which 2 must be raised to get 8.
Since $2^3 = 8$, we have $^2\log 8 = 3$.

For the third term, $^2\log 16$:
We need to find the power to which 2 must be raised to get 16.
Since $2^4 = 16$, we have $^2\log 16 = 4$.

The original expression is $^2\log 4 + ^2\log 8 - ^2\log 16$.
Substitute the values found in Step 1:

$$2 + 3 - 4$$

$$2 + 3 - 4 = 5 - 4 = 1$$

Alternatively, using logarithm properties:
The expression is $^2\log 4 + ^2\log 8 - ^2\log 16$.
Using the properties $\log_b M + \log_b N = \log_b (MN)$ and $\log_b M - \log_b N = \log_b \left(\frac{M}{N}\right)$:

$$^2\log 4 + ^2\log 8 - ^2\log 16 = ^2\log (4 \times 8) - ^2\log 16$$ $$= ^2\log 32 - ^2\log 16$$ $$= ^2\log \left(\frac{32}{16}\right)$$ $$= ^2\log 2$$

Since $2^1 = 2$, $^2\log 2 = 1$.

The calculated value is 1, which corresponds to option D.

The final answer is $\text{D. 1}$.