For positive real numbers a, b, c, the least value of alogb – logc +

1.	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
Answer» (c) 3. a > 0, b > 0, c > 0 ⇒ loga, logb, logc are all defined. Also a^{logb – logc}, b^logc–loga, c^loga–logb are all positive quantities. ∴ Applying AM > GM, we have \(\frac13\)[ a^{logb – logc}+ b^logc–loga+ c^loga–logb] > [a^{logb – logc}. b^logc–loga. c^loga–logb]^\(\frac13\) ...(i) Let x = a^{log b – log c} . b^{log c – log a}. c^{log a – log b} ⇒ log x = (log b – log c) log a + (log c – log a) log b + (log a – log b) log c Now, log_e x = 0 ⇒ x = e⁰ = 1. ∴ (i) ⇒ \(\frac13\)[ a^{logb – logc}+ b^logc–loga+ c^loga–logb] > 1 ⇒ a^{log b – log c} + b^{log c – log a} + c^{log a – log b} > 3 ⇒ The least value of a^{log b – log c} + b^{log c – log a} + c^{log a – log b} is 3.

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

Answer»

(c) 3.

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

Also a^{logb – logc}, b^logc–loga, c^loga–logb are all positive quantities.

∴ Applying AM > GM, we have

\(\frac13\)[ a^{logb – logc}+ b^logc–loga+ c^loga–logb] > [a^{logb – logc}. b^logc–loga. c^loga–logb]^\(\frac13\) ...(i)

Let x = a^{log b – log c} . b^{log c – log a}. c^{log a – log b}

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

Now, log_e x = 0 ⇒ x = e⁰ = 1.

∴ (i) ⇒ \(\frac13\)[ a^{logb – logc}+ b^logc–loga+ c^loga–logb] > 1

⇒ a^{log b – log c} + b^{log c – log a} + c^{log a – log b} > 3

⇒ The least value of a^{log b – log c} + b^{log c – log a} + c^{log a – log b} is 3.