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Express in terms of log 2 and log 3 : ...

Express in terms of log 2 and log 3 :
(i) log 36
(ii) log 144
(iii) log 4.5
(iv) `"log"(26)/(51) - "log" (91)/(119)`
(v) `"log"(75)/(16) - "2 log" (5)/(9) + "log" (32)/(243)`

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To express the given logarithmic expressions in terms of log 2 and log 3, we will use the properties of logarithms. Here is the step-by-step solution for each part of the question: ### (i) log 36 1. **Rewrite 36**: \[ 36 = 6^2 \] Therefore, \[ \log 36 = \log(6^2) \] 2. **Apply the power rule**: \[ \log(6^2) = 2 \cdot \log 6 \] 3. **Rewrite 6 in terms of 2 and 3**: \[ 6 = 2 \cdot 3 \implies \log 6 = \log(2 \cdot 3) = \log 2 + \log 3 \] 4. **Substitute back**: \[ 2 \cdot \log 6 = 2(\log 2 + \log 3) = 2 \log 2 + 2 \log 3 \] Thus, \[ \log 36 = 2 \log 2 + 2 \log 3 \] ### (ii) log 144 1. **Rewrite 144**: \[ 144 = 12^2 \] Therefore, \[ \log 144 = \log(12^2) \] 2. **Apply the power rule**: \[ \log(12^2) = 2 \cdot \log 12 \] 3. **Rewrite 12 in terms of 2 and 3**: \[ 12 = 4 \cdot 3 = 2^2 \cdot 3 \implies \log 12 = \log(2^2 \cdot 3) = \log(2^2) + \log 3 = 2 \log 2 + \log 3 \] 4. **Substitute back**: \[ 2 \cdot \log 12 = 2(2 \log 2 + \log 3) = 4 \log 2 + 2 \log 3 \] Thus, \[ \log 144 = 4 \log 2 + 2 \log 3 \] ### (iii) log 4.5 1. **Rewrite 4.5**: \[ 4.5 = \frac{9}{2} \] Therefore, \[ \log 4.5 = \log\left(\frac{9}{2}\right) \] 2. **Apply the quotient rule**: \[ \log\left(\frac{9}{2}\right) = \log 9 - \log 2 \] 3. **Rewrite 9**: \[ 9 = 3^2 \implies \log 9 = 2 \log 3 \] 4. **Substitute back**: \[ \log 4.5 = 2 \log 3 - \log 2 \] Thus, \[ \log 4.5 = 2 \log 3 - \log 2 \] ### (iv) log(26/51) - log(91/119) 1. **Combine the logarithms**: \[ \log\left(\frac{26}{51}\right) - \log\left(\frac{91}{119}\right) = \log\left(\frac{26}{51} \cdot \frac{119}{91}\right) \] 2. **Simplify the fraction**: \[ = \log\left(\frac{26 \cdot 119}{51 \cdot 91}\right) \] 3. **Factor the numbers**: - \(26 = 2 \cdot 13\) - \(51 = 3 \cdot 17\) - \(91 = 7 \cdot 13\) - \(119 = 7 \cdot 17\) 4. **Substitute and simplify**: \[ = \log\left(\frac{2 \cdot 13 \cdot 119}{3 \cdot 17 \cdot 91}\right) = \log\left(\frac{2}{3}\right) \] 5. **Rewrite in terms of log 2 and log 3**: \[ \log\left(\frac{2}{3}\right) = \log 2 - \log 3 \] Thus, \[ \log\left(\frac{26}{51}\right) - \log\left(\frac{91}{119}\right) = \log 2 - \log 3 \] ### (v) log(75/16) - 2 log(5/9) + log(32/243) 1. **Rewrite each term**: \[ \log\left(\frac{75}{16}\right) - 2 \log\left(\frac{5}{9}\right) + \log\left(\frac{32}{243}\right) \] 2. **Apply properties of logarithms**: \[ = \log\left(\frac{75}{16}\right) - \log\left(\frac{5^2}{9^2}\right) + \log\left(\frac{32}{243}\right) \] 3. **Combine the logarithms**: \[ = \log\left(\frac{75 \cdot 9^2 \cdot 32}{16 \cdot 25}\right) \] 4. **Factor the numbers**: - \(75 = 3 \cdot 5^2\) - \(16 = 2^4\) - \(32 = 2^5\) - \(243 = 3^5\) 5. **Substitute and simplify**: \[ = \log\left(\frac{3 \cdot 5^2 \cdot 9^2 \cdot 2^5}{2^4 \cdot 5^2 \cdot 3^5}\right) = \log\left(\frac{3 \cdot 2^5}{2^4 \cdot 3^5}\right) = \log\left(\frac{2}{3^4}\right) \] 6. **Rewrite in terms of log 2 and log 3**: \[ = \log 2 - 4 \log 3 \] Thus, \[ \log\left(\frac{75}{16}\right) - 2 \log\left(\frac{5}{9}\right) + \log\left(\frac{32}{243}\right) = \log 2 - 4 \log 3 \] ### Summary of Answers: 1. \( \log 36 = 2 \log 2 + 2 \log 3 \) 2. \( \log 144 = 4 \log 2 + 2 \log 3 \) 3. \( \log 4.5 = 2 \log 3 - \log 2 \) 4. \( \log\left(\frac{26}{51}\right) - \log\left(\frac{91}{119}\right) = \log 2 - \log 3 \) 5. \( \log\left(\frac{75}{16}\right) - 2 \log\left(\frac{5}{9}\right) + \log\left(\frac{32}{243}\right) = \log 2 - 4 \log 3 \)
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