log-0-5-2-8-2x-x-2-7log-2-8-2x-x-2-lt-12-

Question Number 85347 by jagoll last updated on 21/Mar/20

\log_{0.5}^{2} (8 + 2 x - x^{2}) - {7 \log}_{2} (8 + 2 x - x^{2}) < - 12

Commented by john santu last updated on 21/Mar/20

(i) 8 + 2 x - x^{2} > 0

x^{2} - 2 x - 8 < 0

(x - 4) (x + 2) < 0

- 2 < x < 4

(ii) \log_{0.5}^{2} (8 + 2 x - x^{2}) = \log_{2}^{2} (8 + 2 x - x^{2})

let \log_{2} (8 + 2 x - x^{2}) = t

\Rightarrow t^{2} - 7 t + 12 < 0

3 < t < 4 \Rightarrow 8 < 8 + 2 x - x^{2} < 16

- 16 < x^{2} - 2 x - 8 < - 8

- 7 < x^{2} - 2 x + 1 < 1

- 7 < {(x - 1)}^{2} < 1 \Rightarrow {(x - 1)}^{2} - 1 < 0

0 < x < 2 \Rightarrow ∴ 0 < x < 2 \leftarrow solution