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Question Number 21357 by Tinkutara last updated on 21/Sep/17

Solve : log_(2x+3) x^2 < 1

$$\mathrm{Solve}\::\:\mathrm{log}_{\mathrm{2}{x}+\mathrm{3}} {x}^{\mathrm{2}} \:<\:\mathrm{1} \\ $$

Answered by dioph last updated on 21/Sep/17

2x + 3 > 0 ⇒ x > −(3/2) 2x + 3 ≠ 1 ⇒ x ≠ −1 ((log x^2 )/(log 2x+3)) < 1 Case 1: x < −1 (2x + 3 < 1) log 2x + 3 < 0 ⇒ log x^2 > log 2x+3 ⇒ log (x^2 /(2x+3)) > 0 ⇒ (x^2 /(2x+3)) > 1 ⇒ x^2 −2x − 3 > 0 ⇒ x ∈ (−∞,−1) ∪ (3, +∞) but x < −1 ⇒ x ∈ (−∞,−1) Case 2: x > −1 (2x+3 > 1) log 2x+3 > 0 ⇒ log x^2 < log 2x+3 ⇒ log (x^2 /(2x+3)) < 0 ⇒ (x^2 /(2x+3)) < 1 ⇒ x^2 −2x−3 < 0 ⇒ x ∈ (−1,3) x > −1 ⇒ x ∈ (−1, 3) putting together: x ∈ (−(3/2), 3) − {−1}

$$\mathrm{2}{x}\:+\:\mathrm{3}\:>\:\mathrm{0}\:\Rightarrow\:{x}\:>\:−\frac{\mathrm{3}}{\mathrm{2}} \\ $$ $$\mathrm{2}{x}\:+\:\mathrm{3}\:\neq\:\mathrm{1}\:\Rightarrow\:{x}\:\neq\:−\mathrm{1} \\ $$ $$\frac{\mathrm{log}\:{x}^{\mathrm{2}} }{\mathrm{log}\:\mathrm{2}{x}+\mathrm{3}}\:<\:\mathrm{1} \\ $$ $$\mathrm{Case}\:\mathrm{1}:\:{x}\:<\:−\mathrm{1}\:\left(\mathrm{2}{x}\:+\:\mathrm{3}\:<\:\mathrm{1}\right) \\ $$ $$\mathrm{log}\:\mathrm{2}{x}\:+\:\mathrm{3}\:<\:\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{log}\:{x}^{\mathrm{2}} \:>\:\mathrm{log}\:\mathrm{2}{x}+\mathrm{3} \\ $$ $$\Rightarrow\:\mathrm{log}\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}\:>\:\mathrm{0} \\ $$ $$\Rightarrow\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}\:>\:\mathrm{1} \\ $$ $$\Rightarrow\:{x}^{\mathrm{2}} \:\:−\mathrm{2}{x}\:−\:\mathrm{3}\:>\:\mathrm{0} \\ $$ $$\Rightarrow\:{x}\:\in\:\left(−\infty,−\mathrm{1}\right)\:\cup\:\left(\mathrm{3},\:+\infty\right) \\ $$ $$\mathrm{but}\:{x}\:<\:−\mathrm{1}\:\Rightarrow\:{x}\:\in\:\left(−\infty,−\mathrm{1}\right) \\ $$ $$\mathrm{Case}\:\mathrm{2}:\:{x}\:>\:−\mathrm{1}\:\left(\mathrm{2}{x}+\mathrm{3}\:>\:\mathrm{1}\right) \\ $$ $$\mathrm{log}\:\mathrm{2}{x}+\mathrm{3}\:>\:\mathrm{0} \\ $$ $$\Rightarrow\:\mathrm{log}\:{x}^{\mathrm{2}} \:<\:\mathrm{log}\:\mathrm{2}{x}+\mathrm{3} \\ $$ $$\Rightarrow\:\mathrm{log}\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}\:<\:\mathrm{0} \\ $$ $$\Rightarrow\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}{x}+\mathrm{3}}\:<\:\mathrm{1} \\ $$ $$\Rightarrow\:{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{3}\:<\:\mathrm{0} \\ $$ $$\Rightarrow\:{x}\:\in\:\left(−\mathrm{1},\mathrm{3}\right) \\ $$ $${x}\:>\:−\mathrm{1}\:\Rightarrow\:{x}\:\in\:\left(−\mathrm{1},\:\mathrm{3}\right) \\ $$ $$\mathrm{putting}\:\mathrm{together}: \\ $$ $${x}\:\in\:\left(−\frac{\mathrm{3}}{\mathrm{2}},\:\mathrm{3}\right)\:−\:\left\{−\mathrm{1}\right\} \\ $$

Commented byTinkutara last updated on 21/Sep/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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