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Question Number 83285 by jagoll last updated on 29/Feb/20

3^((x+2)(x−4)) ≤ 7^(x+2) find solution

$$\mathrm{3}^{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{4}\right)} \:\leqslant\:\mathrm{7}^{\mathrm{x}+\mathrm{2}} \\ $$$$\mathrm{find}\:\mathrm{solution} \\ $$

Commented by jagoll last updated on 29/Feb/20

7^(log_7 (3^((x+2)(x−4)) ) ) ≤ 7^(x+2) ⇒ (x+2)(x−4)log_7 (3) ≤ x+2 ⇒ (x+2) [ (x−4)log_7 (3)−1] ≤ 0 −2 ≤ x ≤ 4+ log_3 (7)

$$\mathrm{7}^{\mathrm{log}_{\mathrm{7}} \:\left(\mathrm{3}^{\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{4}\right)} \right)\:} \leqslant\:\mathrm{7}^{\mathrm{x}+\mathrm{2}} \\ $$$$\Rightarrow\:\left(\mathrm{x}+\mathrm{2}\right)\left(\mathrm{x}−\mathrm{4}\right)\mathrm{log}_{\mathrm{7}} \:\left(\mathrm{3}\right)\:\leqslant\:\mathrm{x}+\mathrm{2} \\ $$$$\Rightarrow\:\left(\mathrm{x}+\mathrm{2}\right)\:\left[\:\left(\mathrm{x}−\mathrm{4}\right)\mathrm{log}_{\mathrm{7}} \:\left(\mathrm{3}\right)−\mathrm{1}\right]\:\leqslant\:\mathrm{0} \\ $$$$−\mathrm{2}\:\leqslant\:\mathrm{x}\:\leqslant\:\mathrm{4}+\:\mathrm{log}_{\mathrm{3}} \left(\mathrm{7}\right) \\ $$$$ \\ $$

Answered by MJS last updated on 01/Mar/20

(x+2)(x−4)ln 3 ≤(x+2)ln 7 (1) x+2<0 ⇔ x<−2 (x−4)ln 3 ≥ln 7 x≥((ln 7)/(ln 3))+4 ⇒ no solution (2) x+2≥0 ⇔ x≥−2 for x=−2: 3^((x+2)(x−4)) =7^(x+2) for x>−2: (x−4)ln 3 ≤ln 7 x≤((ln 7)/(ln 3))+4 ⇒ −2≤x≤((ln 7)/(ln 3))+4

$$\left({x}+\mathrm{2}\right)\left({x}−\mathrm{4}\right)\mathrm{ln}\:\mathrm{3}\:\leqslant\left({x}+\mathrm{2}\right)\mathrm{ln}\:\mathrm{7} \\ $$$$\left(\mathrm{1}\right)\:{x}+\mathrm{2}<\mathrm{0}\:\Leftrightarrow\:{x}<−\mathrm{2} \\ $$$$\left({x}−\mathrm{4}\right)\mathrm{ln}\:\mathrm{3}\:\geqslant\mathrm{ln}\:\mathrm{7} \\ $$$${x}\geqslant\frac{\mathrm{ln}\:\mathrm{7}}{\mathrm{ln}\:\mathrm{3}}+\mathrm{4}\:\Rightarrow\:\mathrm{no}\:\mathrm{solution} \\ $$$$\left(\mathrm{2}\right)\:{x}+\mathrm{2}\geqslant\mathrm{0}\:\Leftrightarrow\:{x}\geqslant−\mathrm{2}\:\mathrm{for}\:{x}=−\mathrm{2}:\:\mathrm{3}^{\left({x}+\mathrm{2}\right)\left({x}−\mathrm{4}\right)} =\mathrm{7}^{{x}+\mathrm{2}} \\ $$$$\mathrm{for}\:{x}>−\mathrm{2}: \\ $$$$\left({x}−\mathrm{4}\right)\mathrm{ln}\:\mathrm{3}\:\leqslant\mathrm{ln}\:\mathrm{7} \\ $$$${x}\leqslant\frac{\mathrm{ln}\:\mathrm{7}}{\mathrm{ln}\:\mathrm{3}}+\mathrm{4}\:\Rightarrow\:−\mathrm{2}\leqslant{x}\leqslant\frac{\mathrm{ln}\:\mathrm{7}}{\mathrm{ln}\:\mathrm{3}}+\mathrm{4} \\ $$

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