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Algebra

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4 Nov 2008 12:32:45 IST

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858

Log Inequality

None

Prove that

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Gaurav |spideyunlimited| Ragtah

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4 Nov 2008 12:58:56 IST

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log₇ 10 = 1 + log₇ (10/7) = 1 + log₇ 1.428

log₁₁ 13 = 1 + log₁₁ (13/11) = 1 + log₁₁ 1.18

Let A = log₇ 1.428

and B = log₁₁ 1.18

So, 7^A = 1.428

and 11^B = 1.18

Clearly A and B will both be fractional, and B will be smaller than A; (Since, in case of A, 7 is being reduced to 1.428 but with B, a larger number 11 is being reduced to a correspondingly less value than for A. )

Thus, 1 + log₇ 1.428 > 1 + log₁₁ 1.18

$\log _7 10 > \log_{11} 13$

eunikorn

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Joined: 14 Oct 2008

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5 Nov 2008 10:31:33 IST

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log₇10 > log₁₁13

=> log ₇10 > log₇13 / log₇11

=> log₇10 * log₇11 >log₇13 (log₇11 > 0)

=> log₇(10¹¹) > log ₇13

=> 10¹¹> 13 (base ie. 7 >1)

which is quite obvious !

Avirup Dasgupta

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Joined: 30 Jul 2007

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5 Nov 2008 11:15:16 IST

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For Eunikorn:-

Gaurav |spideyunlimited| Ragtah

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Joined: 16 Dec 2006

Posts: 3373

5 Nov 2008 11:33:49 IST

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ya that's what i have asked him in his nudgebook.

Soumik

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6 Nov 2008 12:45:53 IST

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Great Unikorn! Earned 10 points for a wrong solution!!!

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