Question Number 92319 by jagoll last updated on 06/May/20
$$\mathrm{log}\:_{\mathrm{9}} \left(\mathrm{x}+\frac{\mathrm{7}}{\mathrm{2}}\right).\mathrm{log}\:_{\mathrm{3}/\mathrm{4}} \left(\mathrm{x}^{\mathrm{2}} \right)\:\geqslant\: \\ $$$$\mathrm{log}\:_{\mathrm{3}/\mathrm{4}} \left(\mathrm{x}+\frac{\mathrm{7}}{\mathrm{2}}\right)\: \\ $$
Commented by john santu last updated on 06/May/20
![{ ((x+(7/2) > 0 ⇒ x > −(7/2))),((x^2 > 0 ⇒ x ≠ 0)) :} (1/2)log _3 (x+(7/2)).((log _3 (x^2 ))/(log _3 ((3/4)))) ≥ ((log _3 (x+(7/2)))/(log _3 ((3/4)))) log _3 (x+(7/2)) ((1/2)log _3 (x^2 )−1) ≤ 0 (−(7/2),−3] ∪ [−(5/2),0 ) ∪ (0,3)](https://www.tinkutara.com/question/Q92320.png)
$$\begin{cases}{{x}+\frac{\mathrm{7}}{\mathrm{2}}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:−\frac{\mathrm{7}}{\mathrm{2}}}\\{{x}^{\mathrm{2}} \:>\:\mathrm{0}\:\Rightarrow\:{x}\:\neq\:\mathrm{0}}\end{cases} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:_{\mathrm{3}} \left({x}+\frac{\mathrm{7}}{\mathrm{2}}\right).\frac{\mathrm{log}\:_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)}{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)}\:\geqslant\:\frac{\mathrm{log}\:_{\mathrm{3}} \left({x}+\frac{\mathrm{7}}{\mathrm{2}}\right)}{\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{3}}{\mathrm{4}}\right)} \\ $$$$\mathrm{log}\:_{\mathrm{3}} \left({x}+\frac{\mathrm{7}}{\mathrm{2}}\right)\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:_{\mathrm{3}} \left({x}^{\mathrm{2}} \right)−\mathrm{1}\right)\:\leqslant\:\mathrm{0} \\ $$$$\left(−\frac{\mathrm{7}}{\mathrm{2}},−\mathrm{3}\right]\:\cup\:\left[−\frac{\mathrm{5}}{\mathrm{2}},\mathrm{0}\:\right)\:\cup\:\left(\mathrm{0},\mathrm{3}\right)\: \\ $$
Commented by jagoll last updated on 06/May/20
Is this the cut step? why so short?
Commented by jagoll last updated on 06/May/20
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