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Question 112 of 128

Q.Prove that log⁡a+log⁡a2+log⁡a3+…+log⁡an=n(n+1)2log⁡a\log a + \log a^2 + \log a^3 + \ldots + \log a^n = \dfrac{n(n+1)}{2}\log a.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2023Subjective· 2mImportance★★★★★
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Using log⁡ak=klog⁡a\log a^k = k\log a for each term and summing 1+2+⋯+n=n(n+1)21+2+\cdots+n=\dfrac{n(n+1)}{2} proves the identity directly.

Each term log⁡ak=klog⁡a\log a^k = k\log a by the power rule of logarithms. So:

log⁡a+log⁡a2+log⁡a3+⋯+log⁡an=1log⁡a+2log⁡a+3log⁡a+⋯+nlog⁡a\log a + \log a^2 + \log a^3 + \cdots + \log a^n = 1\log a + 2\log a + 3\log a + \cdots + n\log a

=(1+2+3+⋯+n)log⁡a= (1+2+3+\cdots+n)\log a

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