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MISCELLANEOUS EXERCISE 6 (II) · Q169

Q.Simplify log⁡102845−log⁡1035324+log⁡10325432−log⁡101315\log_{10}\dfrac{28}{45}-\log_{10}\dfrac{35}{324}+\log_{10}\dfrac{325}{432}-\log_{10}\dfrac{13}{15}.

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log⁡102845−log⁡1035324+log⁡10325432−log⁡101315=log⁡10[2845×32435×325432×1513]\log_{10}\dfrac{28}{45}-\log_{10}\dfrac{35}{324}+\log_{10}\dfrac{325}{432}-\log_{10}\dfrac{13}{15}=\log_{10}\left[\dfrac{28}{45}\times\dfrac{324}{35}\times\dfrac{325}{432}\times\dfrac{15}{13}\right]

(subtracting a log = adding the log of the reciprocal, then combining all four via the product rule).

Factor and cancel: 28=4⋅728=4\cdot7, 35=5⋅735=5\cdot7 (the 77s cancel); 325=25⋅13325=25\cdot13, and the 1313 in the last fraction cancels the 1313 from 325325. Working through the remaining factors of 2,3,52,3,5 step by step: …

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