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Q.Prove that log⁡7516−2log⁡59+log⁡32243=log⁡2\log\dfrac{75}{16}-2\log\dfrac{5}{9}+\log\dfrac{32}{243}=\log 2 OR By the principle of mathematical induction, prove that, for all integers n≥1n\ge 1, 12+22+32+⋯+n2=n(n+1)(2n+1)61^2+2^2+3^2+\cdots+n^2=\dfrac{n(n+1)(2n+1)}{6}

Puducherry TnboardTamil Nadu HSC First Year (DGE) Board 2026Subjective· 5mImportance★★★★★
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Using 2log⁡(5/9)=log⁡(25/81)2\log(5/9)=\log(25/81) and combining all terms into one logarithm, the fraction inside simplifies exactly to 2, giving log⁡2\log2.

log⁡7516−2log⁡59+log⁡32243\log\dfrac{75}{16}-2\log\dfrac{5}{9}+\log\dfrac{32}{243}

Using the power rule klog⁡a=log⁡akk\log a=\log a^k: 2log⁡59=log⁡(59)2=log⁡25812\log\dfrac{5}{9}=\log\left(\dfrac{5}{9}\right)^2=\log\dfrac{25}{81}

So the expression becomes:

log⁡7516−log⁡2581+log⁡32243\log\dfrac{75}{16}-\log\dfrac{25}{81}+\log\dfrac{32}{243}

Using log⁡a−log⁡b=log⁡(a/b)\log a-\log b=\log(a/b) and log⁡a+log⁡b=log⁡(ab)\log a+\log b=\log(ab), combine into a single logarithm:

=log⁡(7516×8125×32243)=\log\left(\dfrac{75}{16}\times\dfrac{81}{25}\times\dfrac{32}{243}\right)

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