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Q.Prove that log⁡(1+2+3)=log⁡1+log⁡2+log⁡3\log(1+2+3) = \log 1 + \log 2 + \log 3

Meghalaya MboseMBOSE Meghalaya 11th Board 2018Subjective· 2mImportance★★★★★
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Both sides simplify to log⁡6\log 6, since 1+2+3=61+2+3=6 and 1×2×3=61\times2\times3=6 as well.

Using the product rule of logarithms, log⁡a+log⁡b+log⁡c=log⁡(abc)\log a+\log b+\log c=\log(abc).

Right-hand side:

log⁡1+log⁡2+log⁡3=log⁡(1×2×3)=log⁡6\log 1+\log 2+\log 3=\log(1\times2\times3)=\log 6

Left-hand side:

1+2+3=6 ⇒ log⁡(1+2+3)=log⁡61+2+3=6\ \Rightarrow\ \log(1+2+3)=\log 6

Since both sides equal log⁡6\log 6, we have

log⁡(1+2+3)=log⁡1+log⁡2+log⁡3\log(1+2+3)=\log 1+\log 2+\log 3

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