https://file.aiquickdraw.com/mathbot/file/1726792648695shzyybr8.jpg

The image contains two main questions involving surds and logarithmic expressions.

1. **Surd Simplifications**:
   - a. \( 2\sqrt{3} + 3\sqrt{3} - \sqrt{18} \)
   - b. \( 3\sqrt{5} + \sqrt{20} - \frac{1}{3}\sqrt{45} \)
   - c. \( \frac{3}{2\sqrt{6}} \)
   - d. \( \frac{4}{\sqrt{2} - \sqrt{3}} \)

2. **Logarithmic Simplifications**:
   - a. \( 2\log 50 + 2\log 8 - 2\log 100 \)
   - b. \( \log 24 - 2\log 2 + \log \frac{1}{6} \)

Let me solve each part for you.

---

### 1. Surd Simplifications:

**a.**  
\[
2\sqrt{3} + 3\sqrt{3} - \sqrt{18}
\]
First, simplify \( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \).  
Now, the expression becomes:
\[
(2 + 3)\sqrt{3} - 3\sqrt{2} = 5\sqrt{3} - 3\sqrt{2}
\]
This is already in simplified form.

**b.**  
\[
3\sqrt{5} + \sqrt{20} - \frac{1}{3}\sqrt{45}
\]
First, simplify \( \sqrt{20} = 2\sqrt{5} \) and \( \sqrt{45} = 3\sqrt{5} \).  
The expression becomes:
\[
3\sqrt{5} + 2\sqrt{5} - \frac{1}{3} \times 3\sqrt{5} = 3\sqrt{5} + 2\sqrt{5} - \sqrt{5} 
\]
Now simplify the terms:
\[
(3 + 2 - 1)\sqrt{5} = 4\sqrt{5}
\]

**c.**  
\[
\frac{3}{2\sqrt{6}}
\]
To rationalize the denominator, multiply by \( \frac{\sqrt{6}}{\sqrt{6}} \):
\[
\frac{3}{2\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{3\sqrt{6}}{12} = \frac{\sqrt{6}}{4}
\]

**d.**  
\[
\frac{4}{\sqrt{2} - \sqrt{3}}
\]
To rationalize the denominator, multiply by the conjugate \( \sqrt{2} + \sqrt{3} \):
\[
\frac{4}{\sqrt{2} - \sqrt{3}} \times \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} = \frac{4(\sqrt{2} + \sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{4(\sqrt{2} + \sqrt{3})}{2 - 3} = \frac{4(\sqrt{2} + \sqrt{3})}{-1}
\]
Thus, the result is:
\[
-4(\sqrt{2} + \sqrt{3})
\]

---

### 2. Logarithmic Simplifications:

**a.**  
\[
2\log 50 + 2\log 8 - 2\log 100
\]
Using the power rule of logarithms, \( n\log a = \log a^n \):
\[
\log 50^2 + \log 8^2 - \log 100^2 = \log 2500 + \log 64 - \log 10000
\]
Now, apply the logarithmic addition and subtraction rules:
\[
\log \left( \frac{2500 \times 64}{10000} \right) = \log \left( \frac{160000}{10000} \right) = \log 16
\]

**b.**  
\[
\log 24 - 2\log 2 + \log \frac{1}{6}
\]
Using the power rule and simplifying:
\[
\log 24 - \log 2^2 + \log \frac{1}{6} = \log 24 - \log 4 + \log \frac{1}{6}
\]
Apply logarithmic addition and subtraction:
\[
\log \left( \frac{24}{4} \times \frac{1}{6} \right) = \log \left( \frac{6}{6} \right) = \log 1 = 0
\]

---

Do you have any questions, or would you like further details on any specific steps?

### Related Questions:
1. How do you rationalize denominators involving surds?
2. What are the common logarithmic properties and how are they applied?
3. How can you simplify square roots of non-perfect squares?
4. Can surds always be simplified, or are there cases where they cannot?
5. How is the change of base formula applied in logarithmic problems?

**Tip:** When dealing with logarithms, always check for opportunities to apply the power, product, and quotient rules to simplify expressions easily.

Simplify the following surds and logarithmic expressions:

1. Surd Simplifications:
   a. 2√3 + 3√3 - √18
   b. 3√5 + √20 - (1/3)√45
   c. 3/(2√6)
   d. 4/(√2 - √3)

2. Logarithmic Simplifications:
   a. 2 log 50 + 2 log 8 - 2 log 100
   b. log 24 - 2 log 2 + log (1/6)

Learn how to simplify surds and logarithmic expressions with detailed step-by-step solutions. This problem covers surd rationalization, logarithmic properties, and simplification techniques. Suitable for students in Grades 9-12.

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