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Exercise 2.12 · Q7

Q.Prove that log⁡2+16log⁡1615+12log⁡2524+7log⁡8180=1\log 2+16\log\dfrac{16}{15}+12\log\dfrac{25}{24}+7\log\dfrac{81}{80}=1.

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Step 1. Expand each log using the quotient rule and prime factorisations (16=24,15=3⋅5,25=52,24=23⋅3,81=34,80=24⋅516=2^4,15=3\cdot5,25=5^2,24=2^3\cdot3,81=3^4,80=2^4\cdot5):

16log⁡1615=16[4log⁡2−log⁡3−log⁡5]=64log⁡2−16log⁡3−16log⁡516\log\frac{16}{15}=16[4\log2-\log3-\log5]=64\log2-16\log3-16\log5.

12log⁡2524=12[2log⁡5−3log⁡2−log⁡3]=−36log⁡2−12log⁡3+24log⁡512\log\frac{25}{24}=12[2\log5-3\log2-\log3]=-36\log2-12\log3+24\log5.

7log⁡8180=7[4log⁡3−4log⁡2−log⁡5]=−28log⁡2+28log⁡3−7log⁡57\log\frac{81}{80}=7[4\log3-4\log2-\log5]=-28\log2+28\log3-7\log5.

Step 2. Collect the coefficient of log⁡2\log2 (including the initial log⁡2\log2 term): 1+64−36−28=11+64-36-28=1. …

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