To solve the expression $\log_2(16) + \log\left(\frac{1}{1000}\right)$, let's break it down into two parts:

1. Calculate $\log_2(16)$:

• The base-2 logarithm of 16 can be expressed as: $\log_2(16) = \log_2(2^4)$
• Since $\log_b(a^c) = c \cdot \log_b(a)$, we get: $\log_2(16) = 4 \cdot \log_2(2) = 4 \cdot 1 = 4$

2. Calculate $\log\left(\frac{1}{1000}\right)$:

• The logarithm of a fraction can be calculated as: $\log\left(\frac{1}{1000}\right) = \log\left(10^{-3}\right)$
• Using the property $\log(a^b) = b \cdot \log(a)$, we have: $\log\left(10^{-3}\right) = -3 \cdot \log(10)$
• Since $\log(10)$ in base 10 is 1, we get: $\log\left(\frac{1}{1000}\right) = -3 \cdot 1 = -3$

3. Add the two results together:

$\log_2(16) + \log\left(\frac{1}{1000}\right) = 4 + (-3) = 1$

So, the solution is $\boxed{1}$.

Would you like to dive deeper into logarithmic properties, or do you have any other questions?

Here are 5 related questions to explore further:

1. How would you solve $\log_2(32)$?
2. What is the value of $\log\left(\frac{1}{100}\right)$?
3. How do logarithm properties apply to solve $\log_5(25) + \log(100)$?
4. Can you evaluate $\log_2(1)$ and explain why it results in that value?
5. What is the inverse function of a logarithm, and how is it used?

Tip: Always remember that $\log_b(1) = 0$ for any base $b$ because any number raised to the power of zero is 1.