Exercise 2.3 Simplify the following logarithmic expressions without using a calculator: b) \( \log _{3}\left(\log _{5} 125\right) \) a. \( \frac{\log 343}{\log 49} \) d) \( \left(\log _{2} 128\right)\left(\log _{4} 256\right) \) a) \( 2 \log _{3} 2+\log _{3} 10+2 \log _{3} 3-\log _{3} 40 \) f) \( \log _{2} 8^{2}-\log _{7} 49+\log _{5} 1 \) e) \( \frac{\log 4+\log 25}{\log 0,01} \) b) \( \log _{6} 36+\log \frac{1}{1000}-\log _{3} 81 \) g) \( 3 \log _{2} 4+\log _{2} 80-\log _{2} 5 \) i) \( \log _{4} 6+2 \log _{4} 24-3 \log _{4} 3-3 \log _{4} 2 \) y) \( \frac{\log 25}{\log 0,2} \) k) \( \log _{3} \frac{1}{81} \div \log _{3} 9 \) 1) \( \frac{\log _{4} 64-\log _{4} 16}{\log _{4} 256-\log _{4} 1} \) m) \( \frac{\log 216}{\log 36}+2 \log _{3} \sqrt{9} \) 12) \( \frac{\log 500-\log 5}{\log 0,001} \) \( \frac{\log _{x} 243}{\log _{x} \frac{1}{27}} \) p) \( \log _{2}(28+4)-\log _{2} 128 \)

We wish to simplify each expression without a calculator. (Unless a base is indicated, “log” means common logarithm [base 10].) One acceptable set of answers is given below. (Any answer equivalent to these is correct.) ────────────────────────────── 1. b) log₃(log₅ 125) Step 1. Compute log₅ 125. Notice 125 = 5³ so   log₅ 125 = 3. Step 2. Then the expression becomes log₃ 3, and since 3 = 3¹, we have   log₃ 3 = 1. Answer: 1 ────────────────────────────── 2. a) [log 343]⁄[log 49] Write 343 and 49 in terms of 7:   343 = 7³ and 49 = 7². Thus,   (log 343)/(log 49) = (3·log 7)/(2·log 7) = 3/2. Answer: 3/2 ────────────────────────────── 3. d) (log₂ 128)·(log₄ 256) Evaluate each logarithm:   128 = 2⁷ ⟹ log₂ 128 = 7.   256 = 4⁴ (since 4⁴ = 256) ⟹ log₄ 256 = 4. Then, the product is 7·4 = 28. Answer: 28 ────────────────────────────── 4. a) 2·log₃ 2 + log₃ 10 + 2·log₃ 3 – log₃ 40 Combine using logarithm laws: • 2·log₃ 2 = log₃ (2²) = log₃ 4. • 2·log₃ 3 = log₃ (3²) = log₃ 9. Thus, the expression becomes   log₃ 4 + log₃ 10 + log₃ 9 – log₃ 40. Combine the sums and difference:   = log₃ [(4·10·9)/40]. Compute the inside:   4·10·9 = 360 and 360/40 = 9. So, we have log₃ 9. Since 9 = 3², this equals 2. Answer: 2 ────────────────────────────── 5. f) log₂(8²) – log₇ 49 + log₅ 1 Work term‐by‐term: • 8² = 64, and since 64 = 2⁶, log₂ 64 = 6. • 49 = 7²,      log₇ 49 = 2. • log₅ 1 = 0   (because any log of 1 is 0). Thus, the result is 6 – 2 + 0 = 4. Answer: 4 ────────────────────────────── 6. e) (log 4 + log 25)⁄(log 0.01) Combine numerator using log rules:   log 4 + log 25 = log (4·25) = log 100. Since log 100 = 2 (base 10), and   0.01 = 10^(–2) ⟹ log 0.01 = –2, the expression becomes   2/(–2) = –1. Answer: –1 ────────────────────────────── 7. b) log₆ 36 + log (1/1000) – log₃ 81 Evaluate each piece: • 36 = 6² so   log₆ 36 = 2. • 1/1000 = 10^(–3) so log (1/1000) = –3. • 81 = 3⁴ so   log₃ 81 = 4. Thus, the overall expression is 2 + (–3) – 4 = –5. Answer: –5 ────────────────────────────── 8. g) 3·log₂ 4 + log₂ 80 – log₂ 5 First, write 3·log₂ 4 as a single logarithm:   3·log₂ 4 = log₂ (4³) = log₂ 64. Then combine:   log₂ 64 + log₂ 80 – log₂ 5 = log₂ [(64·80)/5]. Since 80⁄5 = 16, we have log₂ (64·16) = log₂ 1024. Since 1024 = 2¹⁰, the answer is 10. Answer: 10 ────────────────────────────── 9. i) log₄ 6 + 2·log₄ 24 – 3·log₄ 3 – 3·log₄ 2 Write each term as a single log:   2·log₄ 24 = log₄ (24²). Also, combine all terms:   = log₄ [6·(24²)⁄(3³·2³)]. Compute the