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MISCELLANEOUS EXERCISE 6 (II) · Q164

Q.For any base show that log⁡(1+2+3)=log⁡1+log⁡2+log⁡3\log(1+2+3)=\log1+\log2+\log3.

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log⁡(1+2+3)=log⁡6\log(1+2+3)=\log6 (just adding the numbers first).

log⁡1+log⁡2+log⁡3=0+log⁡2+log⁡3=log⁡(2×3)=log⁡6\log1+\log2+\log3=0+\log2+\log3=\log(2\times3)=\log6 (using log⁡1=0\log1=0 for any base, then the product rule). …

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