Let's go through the questions in the image one by one and solve them:

a) Convert into logarithmic form $5^3 = 125$

In logarithmic form, $a^b = c$ is written as $\log_a{c} = b$.

Thus: $\log_5{125} = 3$

b) Determine the value of $\log{\sqrt{729}}$

First, simplify $\sqrt{729}$. Since $729 = 27^2$, $\sqrt{729} = 27$.

Now find $\log{27}$. Assuming it’s the common logarithm (base 10): $\log{27} \approx 1.431$

c) $\log{1} + 2\log{2} + 3\log{\sqrt{2}}$

We can simplify each term:

1. $\log{1} = 0$ (since the logarithm of 1 in any base is 0).
2. $2\log{2} = \log{2^2} = \log{4}$.
3. $\log{\sqrt{2}} = \frac{1}{2}\log{2}$, so $3\log{\sqrt{2}} = 3 \times \frac{1}{2} \log{2} = \frac{3}{2} \log{2}$.

Thus: $\log{1} + 2\log{2} + 3\log{\sqrt{2}} = \log{4} + \frac{3}{2} \log{2}$

d) $\log{27} + 3\log{\sqrt{81}}$

We simplify each term:

1. $\log{27} = \log{3^3} = 3\log{3}$.
2. $\log{\sqrt{81}} = \log{9}$, and $9 = 3^2$, so $\log{9} = 2\log{3}$. Hence $3\log{\sqrt{81}} = 3 \times 2 \log{3} = 6\log{3}$.

Thus: $\log{27} + 3\log{\sqrt{81}} = 3\log{3} + 6\log{3} = 9\log{3}$

e) $\log{0.5} + \log{18}$

Using the logarithm product rule, $\log{a} + \log{b} = \log{(a \times b)}$:

$\log{0.5} + \log{18} = \log{(0.5 \times 18)} = \log{9}$

f) $\log{\sqrt{8}} + 2\log{5} - \log{75} + \log{18} - \log{8}$

We simplify using properties of logarithms:

1. $\log{\sqrt{8}} = \frac{1}{2} \log{8}$.
2. $2\log{5} = \log{5^2} = \log{25}$.
3. $\log{75} = \log{(3 \times 25)} = \log{3} + \log{25}$.

Now combine terms using the product and quotient rules of logarithms.

g) $\log{7} = \frac{\log{x}}{\log{y}}$

This is simply a logarithmic equation where $\log{7}$ is equal to the quotient of $\log{x}$ and $\log{y}$. Without further context, we cannot solve for $x$ and $y$.

h) $\frac{\log{9}}{\log{3}}$

Using logarithmic properties, $\log{9} = \log{3^2} = 2\log{3}$. Hence:

$\frac{\log{9}}{\log{3}} = \frac{2\log{3}}{\log{3}} = 2$

i) $\log{7} \times 7 \log{125}$

We simplify $\log{125} = \log{5^3} = 3\log{5}$, so:

$7 \log{125} = 7 \times 3\log{5} = 21\log{5}$

Thus the expression becomes $\log{7} \times 21\log{5}$.

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

Relative Questions:

1. How are logarithmic properties applied to solve expressions involving multiple terms?
2. Can you explain the relationship between logarithms and exponents?
3. How do the rules for products and quotients in logarithms simplify complex expressions?
4. Why is $\log{1} = 0$ for any base?
5. What are the implications of changing the logarithmic base in calculations?

Tip:

Always remember to simplify radicals and exponents before applying logarithmic properties for easier calculations.