Multiply $ \color{blue}{ln} $ by $ \left( x-2\right) $ $$ \color{blue}{ln} \cdot \left( x-2\right) = lnx-2ln $$

Multiply $4$ by $ \dfrac{x}{24} $ to get $ \dfrac{ 4x }{ 24 } $.

Step 1: Write $ 4 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 4 \cdot \frac{x}{24} & \xlongequal{\text{Step 1}} \frac{4}{\color{red}{1}} \cdot \frac{x}{24} \xlongequal{\text{Step 2}} \frac{ 4 \cdot x }{ 1 \cdot 24 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4x }{ 24 } \end{aligned} $$

Multiply $ln$ by $ \dfrac{4x}{24} $ to get $ \dfrac{ 4lnx }{ 24 } $.

Step 1: Write $ ln $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} ln \cdot \frac{4x}{24} & \xlongequal{\text{Step 1}} \frac{ln}{\color{red}{1}} \cdot \frac{4x}{24} \xlongequal{\text{Step 2}} \frac{ ln \cdot 4x }{ 1 \cdot 24 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 4lnx }{ 24 } \end{aligned} $$

Subtract $lnx-2ln$ from $ \dfrac{4lnx}{24} $ to get $ \dfrac{ \color{purple}{ -20lnx+48ln } }{ 24 }$.

Step 1: Write $ lnx-2ln $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{24}$.

$$ \begin{aligned} \frac{4lnx}{24} -lnx-2ln & \xlongequal{\text{Step 1}} \frac{4lnx}{24} - \frac{lnx-2ln}{\color{red}{1}} = \frac{ 4lnx }{ 24 } - \frac{ \left( lnx-2ln \right) \cdot \color{blue}{ 24 }}{ 1 \cdot \color{blue}{ 24 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 4lnx } }{ 24 } - \frac{ \color{purple}{ 24lnx-48ln } }{ 24 }=\frac{ \color{purple}{ 4lnx - \left( 24lnx-48ln \right) } }{ 24 } = \\[1ex] &=\frac{ \color{purple}{ -20lnx+48ln } }{ 24 } \end{aligned} $$