Let's break down each expression step by step using logarithmic properties. The following key logarithmic rules will be used:

1. Product Rule: $\log(ab) = \log(a) + \log(b)$
2. Quotient Rule: $\log\left(\frac{a}{b}\right) = \log(a) - \log(b)$
3. Power Rule: $\log(a^n) = n\log(a)$

a. $\log(xy^2)$

Using the product rule and the power rule: $\log(xy^2) = \log(x) + \log(y^2)$ $= \log(x) + 2\log(y)$

b. $\log\left(\frac{xy}{z}\right)$

Using the product rule first and then applying the quotient rule: $\log\left(\frac{xy}{z}\right) = \log(xy) - \log(z)$ Now apply the product rule to $\log(xy)$: $= \log(x) + \log(y) - \log(z)$

c. $\log\left(\frac{1}{xyz}\right)$

We can rewrite this as $\log\left(1\right) - \log(xyz)$, and since $\log(1) = 0$, we only have: $\log\left(\frac{1}{xyz}\right) = -\log(xyz)$ Now apply the product rule: $= -(\log(x) + \log(y) + \log(z))$ $= -\log(x) - \log(y) - \log(z)$

d. $\log\left(\sqrt{\frac{x}{yz}}\right)$

First, rewrite the square root as a power of 1/2: $\log\left(\sqrt{\frac{x}{yz}}\right) = \log\left(\left(\frac{x}{yz}\right)^{1/2}\right)$ Now apply the power rule: $= \frac{1}{2}\log\left(\frac{x}{yz}\right)$ Next, apply the quotient rule: $= \frac{1}{2}(\log(x) - \log(yz))$ Finally, apply the product rule to $\log(yz)$: $= \frac{1}{2}(\log(x) - (\log(y) + \log(z)))$ $= \frac{1}{2}(\log(x) - \log(y) - \log(z))$

Final Answers:

a. $\log(xy^2) = \log(x) + 2\log(y)$
b. $\log\left(\frac{xy}{z}\right) = \log(x) + \log(y) - \log(z)$
c. $\log\left(\frac{1}{xyz}\right) = -\log(x) - \log(y) - \log(z)$
d. $\log\left(\sqrt{\frac{x}{yz}}\right) = \frac{1}{2}(\log(x) - \log(y) - \log(z))$

Would you like more details on any of these steps or have any questions?

5 Related Questions:

1. How do you simplify logarithmic expressions involving multiple powers?
2. What are the basic properties of logarithms?
3. How do you expand logarithms using the product, quotient, and power rules?
4. How can logarithmic equations be solved for a specific variable?
5. What is the relationship between logarithms and exponential functions?

Tip:

Always remember that $\log(1) = 0$ for any logarithmic base, which simplifies expressions involving $\log(1)$ quickly!