1. **Problem (a):** Prove without a calculator that $$\log_{10}\left(\frac{170}{7}\right) - \log_{10}\left(\frac{18}{35}\right) + \log_{10}\left(\frac{36}{17}\right) = 2.$$ 2. **Formula and rules:** Recall the logarithm properties: - $$\log a - \log b = \log \left(\frac{a}{b}\right)$$ - $$\log a + \log b = \log (ab)$$ These allow us to combine and simplify logarithmic expressions. 3. **Step-by-step simplification:** \[\begin{aligned} &\log_{10}\left(\frac{170}{7}\right) - \log_{10}\left(\frac{18}{35}\right) + \log_{10}\left(\frac{36}{17}\right) \\ &= \log_{10}\left(\frac{170}{7}\right) + \log_{10}\left(\frac{36}{17}\right) - \log_{10}\left(\frac{18}{35}\right) \\ &= \log_{10}\left(\frac{170}{7} \times \frac{36}{17}\right) - \log_{10}\left(\frac{18}{35}\right) \\ &= \log_{10}\left(\frac{170 \times 36}{7 \times 17}\right) - \log_{10}\left(\frac{18}{35}\right) \\ &= \log_{10}\left(\frac{170 \times 36}{7 \times 17} \times \frac{35}{18}\right) \\ &= \log_{10}\left(\frac{170 \times 36 \times 35}{7 \times 17 \times 18}\right) \end{aligned}\] 4. **Simplify the fraction inside the logarithm:** - Factor and cancel terms: - 170 = 10 \times 17 - 36 = 6 \times 6 - 35 = 7 \times 5 - 18 = 6 \times 3 So, \[\frac{170 \times 36 \times 35}{7 \times 17 \times 18} = \frac{(10 \times 17) \times (6 \times 6) \times (7 \times 5)}{7 \times 17 \times (6 \times 3)}\] Cancel common factors: - Cancel 17 in numerator and denominator - Cancel 7 in numerator and denominator - Cancel one 6 in numerator and denominator Remaining: \[\frac{10 \times 6 \times 5}{3} = \frac{300}{3} = 100\] 5. **Final evaluation:** \[\log_{10}(100) = 2\] Thus, the expression equals 2 as required. --- 6. **Problem (b):** In a geometric sequence with positive terms, the sum of the first and fourth terms is 70, and the sum of the second and third terms is 60. Find the sequence. 7. **Formula and rules:** - A geometric sequence has terms: $$a, ar, ar^2, ar^3, \ldots$$ where $$a$$ is the first term and $$r$$ is the common ratio. - Given sums: $$a + ar^3 = 70$$ $$ar + ar^2 = 60$$ 8. **Set up equations:** \[\begin{cases} a + ar^3 = 70 \\ ar + ar^2 = 60 \end{cases}\] Factor terms: \[\begin{cases} a(1 + r^3) = 70 \\ a r (1 + r) = 60 \end{cases}\] 9. **Express $$a$$ from the first equation:** $$a = \frac{70}{1 + r^3}$$ 10. **Substitute into second equation:** \[\frac{70}{1 + r^3} \times r (1 + r) = 60\] Simplify: \[70 r \frac{1 + r}{1 + r^3} = 60\] Divide both sides by 10: \[7 r \frac{1 + r}{1 + r^3} = 6\] 11. **Rewrite denominator using factorization:** Recall: $$1 + r^3 = (1 + r)(1 - r + r^2)$$ So, \[7 r \frac{1 + r}{(1 + r)(1 - r + r^2)} = 6 \implies 7 r \frac{1}{1 - r + r^2} = 6\] 12. **Solve for $$r$$:** \[7 r = 6 (1 - r + r^2)\] Expand right side: \[7 r = 6 - 6 r + 6 r^2\] Bring all terms to one side: \[6 r^2 - 13 r + 6 = 0\] 13. **Use quadratic formula:** $$r = \frac{13 \pm \sqrt{(-13)^2 - 4 \times 6 \times 6}}{2 \times 6} = \frac{13 \pm \sqrt{169 - 144}}{12} = \frac{13 \pm 5}{12}$$ 14. **Calculate roots:** - $$r = \frac{13 + 5}{12} = \frac{18}{12} = 1.5$$ - $$r = \frac{13 - 5}{12} = \frac{8}{12} = \frac{2}{3}$$ 15. **Find corresponding $$a$$ values:** For $$r=1.5$$: $$a = \frac{70}{1 + (1.5)^3} = \frac{70}{1 + 3.375} = \frac{70}{4.375} = 16$$ For $$r=\frac{2}{3}$$: $$a = \frac{70}{1 + (\frac{2}{3})^3} = \frac{70}{1 + \frac{8}{27}} = \frac{70}{\frac{35}{27}} = 70 \times \frac{27}{35} = 54$$ 16. **Write sequences:** - For $$r=1.5$$ and $$a=16$$: $$16, 24, 36, 54, \ldots$$ - For $$r=\frac{2}{3}$$ and $$a=54$$: $$54, 36, 24, 16, \ldots$$ Both sequences satisfy the conditions and have positive terms. **Final answers:** (a) $$2$$ (b) The geometric sequences are $$16, 24, 36, 54, \ldots$$ or $$54, 36, 24, 16, \ldots$$