Rewrite in standard linear form: $\frac{dy}{dx} + \frac{y}{x\log x} = 2$

Integrating factor: $e^{\int \frac{1}{x\log x}dx} = e^{\ln(\log x)} = \log x$

Multiply through: $\frac{d}{dx}(y \log x) = 2\log x$

Integrate: $y\log x = 2(x\log x - x) + C$

At $x=1$: $y(1)\cdot 0 = 2(0-1)+C \Rightarrow C=2$

So $y\log x = 2x\log x - 2x + 2$. At $x=e$: $y(1) = 2e - 2e + 2 = \mathbf{2}$

Total cost = (negative cost) + (first 10 prints) + (next 10 prints)

Total $= x + 10 \cdot \frac{3x}{5} + 10 \cdot \frac{x}{5} = 45$

$x + \frac{30x}{5} + \frac{10x}{5} = 45$

$x + 6x + 2x = 45$

$9x = 45$

$x = 5$

Therefore $x = 5$ dollars.

Wait, but the marked answer is A (index 0, value 3). Let me verify: if $x = 3$:

Total $= 3 + 10(\frac{9}{5}) + 10(\frac{3}{5}) = 3 + 18 + 6 = 27 \neq 45$

If $x = 5$: Total $= 5 + 10(3) + 10(1) = 5 + 30 + 10 = 45$ ✓

The mathematically correct answer is $x = 5$ (option D, index 3).

Young's Formula for pediatric dosage calculation:

\[\text{Child's dose} = \frac{\text{Age (years)}}{\text{Age (years)} + 12} \times \text{Adult dose}\]

Calculation:

\[\text{Child's dose} = \frac{6}{6 + 12} \times 300\,\text{mg}\]

\[= \frac{6}{18} \times 300\,\text{mg}\]

\[= \frac{1}{3} \times 300\,\text{mg} = 100\,\text{mg}\]

Other pediatric dose formulas:

BSA-based dosing is most precise in clinical practice. Young's formula is used when weight is unavailable.

The reactions are as follows:

• Acidic Medium: $2[Fe(CN)_{6}]^{4-} + H_{2}O_{2} + 2H^{+} \rightarrow 2[Fe(CN)_{6}]^{3-} + 2H_{2}O$. Here, $H_{2}O_{2}$ is reduced to $H_{2}O$.
• Alkaline Medium: $2[Fe(CN)_{6}]^{3-} + H_{2}O_{2} + 2OH^{-} \rightarrow 2[Fe(CN)_{6}]^{4-} + 2H_{2}O + O_{2}$. Here, $H_{2}O_{2}$ is oxidized to $O_{2}$.

Thus, the products are $H_{2}O$ and $(H_{2}O + O_{2})$.