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in Linear Inequations by (23.5k points)
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For positive real numbers a, b, c, the least value of alogb – logc + blogc – loga + cloga – logb is 

(a) 0 

(b) 1 

(c) 3 

(d) 6

1 Answer

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Best answer

(c) 3.

a > 0, b > 0, c > 0 ⇒ loga, logb, logc are all defined.

Also alogb – logc, blogc–loga, cloga–logb are all positive quantities. 

∴ Applying AM > GM, we have 

\(\frac13\)[ alogb – logc + blogc–loga + cloga–logb > [alogb – logc. blogc–loga. cloga–logb]\(\frac13\)      ...(i) 

Let x = alog b – log c . blog c – log a. clog a – log b 

⇒ log x = (log b – log c) log a + (log c – log a) log b + (log a – log b) log c 

Now, loge x = 0 ⇒ x = e0 = 1. 

∴ (i) ⇒ \(\frac13\)[ alogb – logc + blogc–loga + cloga–logb > 1

⇒ alog b – log c + blog c – log a + clog a – log b >

⇒ The least value of alog b – log c + blog c – log a + clog a – log b is 3.

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