1. **Find the value of $x$ satisfying the equation $4 \log_2 2 + \log_2 x = -4$.** Step 1: Simplify $4 \log_2 2$. Since $\log_2 2 = 1$, we have $4 \times 1 = 4$. Step 2: Substitute back into the equation: $$4 + \log_2 x = -4$$ Step 3: Isolate $\log_2 x$: $$\log_2 x = -4 - 4 = -8$$ Step 4: Convert from logarithmic to exponential form: $$x = 2^{-8} = \frac{1}{2^8} = \frac{1}{256}$$ **Answer:** $x = \frac{1}{256}$. 2. **Mr. Ali's house loan problem:** Given: - House price = 220000 - Down payment = 10% of 220000 = 22000 - Loan amount = 220000 - 22000 = 198000 - Loan term = 20 years - Interest rate = 3% compounded monthly (a) **Find the effective monthly interest rate.** Step 1: Convert annual nominal rate to monthly rate: $$i = \frac{3\%}{12} = 0.25\% = 0.0025$$ Step 2: Calculate effective annual rate (EAR): $$EAR = \left(1 + \frac{0.03}{12}\right)^{12} - 1 = (1.0025)^{12} - 1$$ Step 3: Calculate: $$EAR \approx 1.0304 - 1 = 0.0304 = 3.04\%$$ **Answer:** Effective annual rate is approximately 3.04%. (b) **Determine monthly payment.** Step 1: Number of monthly payments: $$n = 20 \times 12 = 240$$ Step 2: Use the loan payment formula: $$P = \frac{r \times PV}{1 - (1 + r)^{-n}}$$ where $r = 0.0025$, $PV = 198000$. Step 3: Calculate denominator: $$1 - (1 + 0.0025)^{-240} = 1 - (1.0025)^{-240}$$ Step 4: Calculate $(1.0025)^{-240} = \frac{1}{(1.0025)^{240}} \approx \frac{1}{1.8194} = 0.5499$ Step 5: Denominator: $$1 - 0.5499 = 0.4501$$ Step 6: Calculate payment: $$P = \frac{0.0025 \times 198000}{0.4501} = \frac{495}{0.4501} \approx 1099.78$$ **Answer:** Monthly payment is approximately 1099.78. (c) **Total interest paid by Mr. Ali.** Step 1: Total amount paid: $$\text{Total paid} = 1099.78 \times 240 = 263947.20$$ Step 2: Interest paid: $$\text{Interest} = 263947.20 - 198000 = 65947.20$$ **Answer:** Total interest paid is approximately 65947.20. (d) **Total amount paid for the house.** Step 1: Add down payment: $$22000 + 263947.20 = 285947.20$$ **Answer:** Total amount paid is approximately 285947.20. (e) **If he missed first 4 payments, how much to pay on 5th payment to settle all arrears?** Step 1: Calculate the loan balance after 4 missed payments plus interest. Step 2: The balance after $k$ payments missed is: $$B_k = P \times \frac{1 - (1 + r)^{-(n-k)}}{r}$$ Step 3: Calculate balance after 4 missed payments: $$B_4 = 1099.78 \times \frac{1 - (1.0025)^{-(240-4)}}{0.0025}$$ Step 4: Calculate $(1.0025)^{-236} = \frac{1}{(1.0025)^{236}} \approx \frac{1}{1.788} = 0.559$$ Step 5: Calculate numerator: $$1 - 0.559 = 0.441$$ Step 6: Calculate balance: $$B_4 = 1099.78 \times \frac{0.441}{0.0025} = 1099.78 \times 176.4 = 193999.99$$ Step 7: Add 4 missed payments with interest: Each missed payment grows with interest for the months missed: - 1st missed payment grows for 4 months: $1099.78 \times (1.0025)^4 = 1099.78 \times 1.01005 = 1110.77$ - 2nd missed payment grows for 3 months: $1099.78 \times (1.0025)^3 = 1099.78 \times 1.00752 = 1107.00$ - 3rd missed payment grows for 2 months: $1099.78 \times (1.0025)^2 = 1099.78 \times 1.00501 = 1103.32$ - 4th missed payment grows for 1 month: $1099.78 \times (1.0025)^1 = 1099.78 \times 1.0025 = 1102.53$ Step 8: Sum of missed payments with interest: $$1110.77 + 1107.00 + 1103.32 + 1102.53 = 4423.62$$ Step 9: Total amount to pay on 5th payment: $$193999.99 + 4423.62 = 198423.61$$ **Answer:** Mr. Ali must pay approximately 198423.61 on the 5th payment to settle all arrears. 3. **Arithmetic sequence problem:** Given: - Sum of first 12 terms $S_{12} = 54$ - Sum of next 12 terms $S_{24} - S_{12} = 486$ Step 1: Sum of first 24 terms: $$S_{24} = 54 + 486 = 540$$ Step 2: Use formula for sum of arithmetic sequence: $$S_n = \frac{n}{2} [2a + (n-1)d]$$ Step 3: Write equations: $$S_{12} = 6[2a + 11d] = 54 \implies 2a + 11d = 9$$ $$S_{24} = 12[2a + 23d] = 540 \implies 2a + 23d = 45$$ Step 4: Subtract first from second: $$(2a + 23d) - (2a + 11d) = 45 - 9$$ $$12d = 36 \implies d = 3$$ Step 5: Substitute $d=3$ into $2a + 11d = 9$: $$2a + 11 \times 3 = 9$$ $$2a + 33 = 9$$ $$2a = 9 - 33 = -24$$ $$a = -12$$ **Answer:** First term $a = -12$, common difference $d = 3$.