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Question Number 105843 by bobhans last updated on 01/Aug/20

(3x)^(1+log _3 (3x)) > 81x^2 find the solution set

$$\left(\mathrm{3}{x}\right)^{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)} \:>\:\mathrm{81}{x}^{\mathrm{2}} \\ $$ $${find}\:{the}\:{solution}\:{set}\: \\ $$

Answered by bemath last updated on 01/Aug/20

(3x)^(1+log _3 (3x)) > 9(3x)^2 (1)domain x > 0 (2) (1+log _3 (3x))(log _3 (3x))> log _3 (9.(3x)^2 ) ⇒let log _3 (3x) = b (1+b)(b) > 2+2b b^2 +b−2b−2 > 0 b^2 −b−2 > 0 → { ((b < −1)),((b >2)) :} → { ((log _3 (3x) <log _3 ((1/3)))),((log _3 (3x) > log _3 (9))) :} → { ((x < (1/9))),((x > 3)) :} solution set : 0 < x < (1/9) ∪ x>3

$$\left(\mathrm{3}{x}\right)^{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)} \:>\:\mathrm{9}\left(\mathrm{3}{x}\right)^{\mathrm{2}} \\ $$ $$\left(\mathrm{1}\right){domain}\:{x}\:>\:\mathrm{0} \\ $$ $$\left(\mathrm{2}\right)\:\left(\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)\right)\left(\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)\right)>\: \\ $$ $$\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{9}.\left(\mathrm{3}{x}\right)^{\mathrm{2}} \right)\: \\ $$ $$\Rightarrow{let}\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)\:=\:{b}\: \\ $$ $$\left(\mathrm{1}+{b}\right)\left({b}\right)\:>\:\mathrm{2}+\mathrm{2}{b} \\ $$ $${b}^{\mathrm{2}} +{b}−\mathrm{2}{b}−\mathrm{2}\:>\:\mathrm{0} \\ $$ $${b}^{\mathrm{2}} −{b}−\mathrm{2}\:>\:\mathrm{0}\:\rightarrow\begin{cases}{{b}\:<\:−\mathrm{1}}\\{{b}\:>\mathrm{2}}\end{cases} \\ $$ $$\rightarrow\begin{cases}{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)\:<\mathrm{log}\:_{\mathrm{3}} \left(\frac{\mathrm{1}}{\mathrm{3}}\right)}\\{\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)\:>\:\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{9}\right)}\end{cases} \\ $$ $$\rightarrow\begin{cases}{{x}\:<\:\frac{\mathrm{1}}{\mathrm{9}}}\\{{x}\:>\:\mathrm{3}}\end{cases} \\ $$ $${solution}\:{set}\::\:\mathrm{0}\:<\:{x}\:<\:\frac{\mathrm{1}}{\mathrm{9}}\:\cup\:{x}>\mathrm{3} \\ $$

Commented byCoronavirus last updated on 01/Aug/20

clean

Answered by bobhans last updated on 01/Aug/20

(3x)^(1+log _3 (3x)) > (3x)^(log _((3x)) (81x^2 )) (3x−1)(1+log _3 (3x)−log _((3x)) (81x^2 ))>0 (3x−1)(2+log _3 (x)−{((4+2log _3 (x))/(1+log _3 (x)))})>0 (3x−1)(2+log _3 (x))(1−(2/(1+log _3 (x))))>0 (3x−1)(2+log _3 (x))(((log _3 (x)−1)/(1+log _3 (x))))>0 (1) since x >0 then 3x−1 >0 so (((2+log _3 (x))(log _3 (x)−1))/(1+log _3 (x))) >0 let log _3 (x) = υ ⇒ (((2+ϑ)(ϑ−1))/(ϑ+1))>0 → { ((−2<ϑ<−1→(1/9)<x<(1/3))),((ϑ>1→x > 3 )) :}

$$\left(\mathrm{3}{x}\right)^{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)} \:>\:\left(\mathrm{3}{x}\right)^{\mathrm{log}\:_{\left(\mathrm{3}{x}\right)} \:\left(\mathrm{81}{x}^{\mathrm{2}} \right)} \\ $$ $$\left(\mathrm{3}{x}−\mathrm{1}\right)\left(\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left(\mathrm{3}{x}\right)−\mathrm{log}\:_{\left(\mathrm{3}{x}\right)} \left(\mathrm{81}{x}^{\mathrm{2}} \right)\right)>\mathrm{0} \\ $$ $$\left(\mathrm{3}{x}−\mathrm{1}\right)\left(\mathrm{2}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)−\left\{\frac{\mathrm{4}+\mathrm{2log}\:_{\mathrm{3}} \left({x}\right)}{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)}\right\}\right)>\mathrm{0} \\ $$ $$\left(\mathrm{3}{x}−\mathrm{1}\right)\left(\mathrm{2}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)\right)\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)}\right)>\mathrm{0} \\ $$ $$\left(\mathrm{3}{x}−\mathrm{1}\right)\left(\mathrm{2}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)\right)\left(\frac{\mathrm{log}\:_{\mathrm{3}} \left({x}\right)−\mathrm{1}}{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)}\right)>\mathrm{0} \\ $$ $$\left(\mathrm{1}\right)\:{since}\:{x}\:>\mathrm{0}\:{then}\:\mathrm{3}{x}−\mathrm{1}\:>\mathrm{0} \\ $$ $${so}\:\frac{\left(\mathrm{2}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)\right)\left(\mathrm{log}\:_{\mathrm{3}} \left({x}\right)−\mathrm{1}\right)}{\mathrm{1}+\mathrm{log}\:_{\mathrm{3}} \left({x}\right)}\:>\mathrm{0} \\ $$ $${let}\:\mathrm{log}\:_{\mathrm{3}} \left({x}\right)\:=\:\upsilon\:\Rightarrow\:\frac{\left(\mathrm{2}+\vartheta\right)\left(\vartheta−\mathrm{1}\right)}{\vartheta+\mathrm{1}}>\mathrm{0} \\ $$ $$\rightarrow\begin{cases}{−\mathrm{2}<\vartheta<−\mathrm{1}\rightarrow\frac{\mathrm{1}}{\mathrm{9}}<{x}<\frac{\mathrm{1}}{\mathrm{3}}}\\{\vartheta>\mathrm{1}\rightarrow{x}\:>\:\mathrm{3}\:}\end{cases} \\ $$

Answered by 1549442205PVT last updated on 01/Aug/20

We need the condition x>0 Take logarithm both two side of ineqn. by base 3 we get (1+log_3 (3x))log_3 (3x)>log_3 (81x^2 )=log_3 81+log_3 (x^2 ) ⇔(2+log_3 x)(1+log_3 x)>2log_3 x+4 Set log_3 x=y we get (y+2)(1+y)−2y−4>0 ⇔y^2 +y−2>0⇔(y−1)(y+2)>0 ⇔ [((y>1)),((y<−2)) ]⇔ [((log_3 x>1⇔x>3)),((log_3 x<−2⇔0<x<(1/9))) ] Thus,the roots of the given ineqn.is x∈(0;(1/9))∪(3;+∞)

$$\mathrm{We}\:\mathrm{need}\:\mathrm{the}\:\mathrm{condition}\:\mathrm{x}>\mathrm{0} \\ $$ $$\mathrm{Take}\:\mathrm{logarithm}\:\mathrm{both}\:\mathrm{two}\:\mathrm{side}\:\mathrm{of} \\ $$ $$\mathrm{ineqn}.\:\mathrm{by}\:\mathrm{base}\:\mathrm{3}\:\mathrm{we}\:\mathrm{get} \\ $$ $$\left(\mathrm{1}+\mathrm{log}_{\mathrm{3}} \left(\mathrm{3x}\right)\right)\mathrm{log}_{\mathrm{3}} \left(\mathrm{3x}\right)>\mathrm{log}_{\mathrm{3}} \left(\mathrm{81x}^{\mathrm{2}} \right)=\mathrm{log}_{\mathrm{3}} \mathrm{81}+\mathrm{log}_{\mathrm{3}} \left(\mathrm{x}^{\mathrm{2}} \right) \\ $$ $$\Leftrightarrow\left(\mathrm{2}+\mathrm{log}_{\mathrm{3}} \mathrm{x}\right)\left(\mathrm{1}+\mathrm{log}_{\mathrm{3}} \mathrm{x}\right)>\mathrm{2log}_{\mathrm{3}} \mathrm{x}+\mathrm{4} \\ $$ $$\mathrm{Set}\:\mathrm{log}_{\mathrm{3}} \mathrm{x}=\mathrm{y}\:\mathrm{we}\:\mathrm{get} \\ $$ $$\left(\mathrm{y}+\mathrm{2}\right)\left(\mathrm{1}+\mathrm{y}\right)−\mathrm{2y}−\mathrm{4}>\mathrm{0} \\ $$ $$\Leftrightarrow\mathrm{y}^{\mathrm{2}} +\mathrm{y}−\mathrm{2}>\mathrm{0}\Leftrightarrow\left(\mathrm{y}−\mathrm{1}\right)\left(\mathrm{y}+\mathrm{2}\right)>\mathrm{0} \\ $$ $$\Leftrightarrow\begin{bmatrix}{\mathrm{y}>\mathrm{1}}\\{\mathrm{y}<−\mathrm{2}}\end{bmatrix}\Leftrightarrow\begin{bmatrix}{\mathrm{log}_{\mathrm{3}} \mathrm{x}>\mathrm{1}\Leftrightarrow\mathrm{x}>\mathrm{3}}\\{\mathrm{log}_{\mathrm{3}} \mathrm{x}<−\mathrm{2}\Leftrightarrow\mathrm{0}<\mathrm{x}<\frac{\mathrm{1}}{\mathrm{9}}}\end{bmatrix} \\ $$ $$\mathrm{Thus},\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{given}\:\mathrm{ineqn}.\mathrm{is} \\ $$ $$\boldsymbol{\mathrm{x}}\in\left(\mathrm{0};\frac{\mathrm{1}}{\mathrm{9}}\right)\cup\left(\mathrm{3};+\infty\right) \\ $$

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