≥ 0

For to be defined

(i) (x – 8) (2 – x) ≥ 0

(x – 2) (x – 8) ≤ 0 ⇒ 2 ≤ x ≤ 8

Now Let say y = log 0.3 (log 2 5 – log 2 2) = log 0.3 (log 2 5/2)

Let y < 0 (assume) then log 0.3 (log 2 5/2) < 0

⇒ log 2 5/2 > 1 ⇒ log 2 5/2 > ⇒ > 2 (7/10) which is true

So y < 0

so denominator is – ve and numerator is +ve, so inequality is not satisfied,

thus = 0

x = 2, 8 .....(i)

Now 2 x – 3 > 31

⇒ (x – 3) > log 2 31 ⇒ x > 3 + log 2 2 4.9 (approx)

⇒ x > 7.9 ⇒ x = 8

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