1. **Problem Statement:** You start selling an NFT at 0.5 and make a 10% profit per sale. The price increases by 10% each time until it reaches 150. From each profit, you deduct 1.5%. We want to find the total deduction by the time the price reaches 150. 2. **Understanding the problem:** - Initial price $P_0 = 0.5$ - Profit per sale = 10% of current price - Deduction = 1.5% of the profit - Final price $P_n = 150$ 3. **Price progression formula:** Each sale increases the price by 10%, so $$P_{k} = P_0 \times (1.1)^k$$ 4. **Find number of sales $n$ until price reaches 150:** $$150 = 0.5 \times (1.1)^n$$ Divide both sides by 0.5: $$300 = (1.1)^n$$ Take natural logarithm: $$\ln(300) = n \ln(1.1)$$ Solve for $n$: $$n = \frac{\ln(300)}{\ln(1.1)} \approx \frac{5.7038}{0.0953} \approx 59.83$$ Since $n$ must be an integer number of sales, we take $n=59$ (price just below 150) or $n=60$ (price just above 150). 5. **Calculate total deduction:** At each sale $k$, profit is 10% of price $P_{k-1}$: $$\text{profit}_k = 0.1 \times P_{k-1}$$ Deduction per sale: $$\text{deduction}_k = 0.015 \times \text{profit}_k = 0.015 \times 0.1 \times P_{k-1} = 0.0015 \times P_{k-1}$$ 6. **Sum of deductions over $n$ sales:** $$\text{total deduction} = \sum_{k=1}^n 0.0015 \times P_{k-1} = 0.0015 \times \sum_{k=0}^{n-1} P_k$$ 7. **Sum of geometric series:** $$\sum_{k=0}^{n-1} P_k = P_0 \times \frac{(1.1)^n - 1}{1.1 - 1} = 0.5 \times \frac{(1.1)^n - 1}{0.1} = 5 \times ((1.1)^n - 1)$$ 8. **Calculate total deduction:** $$\text{total deduction} = 0.0015 \times 5 \times ((1.1)^n - 1) = 0.0075 \times ((1.1)^n - 1)$$ 9. **Plug in $n=60$:** $$ (1.1)^{60} = e^{60 \ln(1.1)} \approx e^{60 \times 0.0953} = e^{5.718} \approx 304.48$$ 10. **Final total deduction:** $$0.0075 \times (304.48 - 1) = 0.0075 \times 303.48 = 2.2761$$ **Answer:** The total deduction by the time the NFT price reaches 150 is approximately **2.28**.