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Question Number 109775 by ZiYangLee last updated on 25/Aug/20
log_a (3x−4a)+log_a 3x=(2/(log_2 a))+log_a (1−2a), where 0<a<(1/2), find the value of x.
$$\mathrm{log}_{{a}} \left(\mathrm{3}{x}−\mathrm{4}{a}\right)+\mathrm{log}_{{a}} \mathrm{3}{x}=\frac{\mathrm{2}}{\mathrm{log}_{\mathrm{2}} {a}}+\mathrm{log}_{{a}} \left(\mathrm{1}−\mathrm{2}{a}\right),\:\mathrm{where}\:\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{2}}, \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{x}. \\ $$
Answered by bemath last updated on 25/Aug/20
△((♭e)/(math))▽ log _a (3x(3x−4a))= 2log _a (2)+log _a (1−2a) log _a (9x^2 −12ax)= log _a (4(1−2a)) 9x^2 −12ax = 4−8a 9x^2 −12ax+8a−4=0 ; where x > ((4a)/3), 0<a<(1/2) x = ((12a + (√(144a^2 −4.9(8a−4))))/(18)) x=((12a+12(√(a^2 −(2a−1))))/(18)) x=((2a+2(√(a^2 −2a+1)))/3) = ((2a+2(√((a−1)^2 )))/3) x=((2a+2∣a−1∣)/3)=((2a−2(a−1))/3) = (2/3)←the answer note ∣a−1∣ = −(a−1) for 0<a<(1/2)
$$\:\:\:\bigtriangleup\frac{\flat{e}}{{math}}\bigtriangledown \\ $$$$\mathrm{log}\:_{{a}} \left(\mathrm{3}{x}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)\right)=\:\mathrm{2log}\:_{{a}} \left(\mathrm{2}\right)+\mathrm{log}\:_{{a}} \left(\mathrm{1}−\mathrm{2}{a}\right) \\ $$$$\mathrm{log}\:_{{a}} \left(\mathrm{9}{x}^{\mathrm{2}} −\mathrm{12}{ax}\right)=\:\mathrm{log}\:_{{a}} \left(\mathrm{4}\left(\mathrm{1}−\mathrm{2}{a}\right)\right) \\ $$$$\mathrm{9}{x}^{\mathrm{2}} −\mathrm{12}{ax}\:=\:\mathrm{4}−\mathrm{8}{a} \\ $$$$\mathrm{9}{x}^{\mathrm{2}} −\mathrm{12}{ax}+\mathrm{8}{a}−\mathrm{4}=\mathrm{0}\:;\:{where}\:{x}\:>\:\frac{\mathrm{4}{a}}{\mathrm{3}},\:\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${x}\:=\:\frac{\mathrm{12}{a}\:+\:\sqrt{\mathrm{144}{a}^{\mathrm{2}} −\mathrm{4}.\mathrm{9}\left(\mathrm{8}{a}−\mathrm{4}\right)}}{\mathrm{18}} \\ $$$${x}=\frac{\mathrm{12}{a}+\mathrm{12}\sqrt{{a}^{\mathrm{2}} −\left(\mathrm{2}{a}−\mathrm{1}\right)}}{\mathrm{18}} \\ $$$${x}=\frac{\mathrm{2}{a}+\mathrm{2}\sqrt{{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}}}{\mathrm{3}}\:=\:\frac{\mathrm{2}{a}+\mathrm{2}\sqrt{\left({a}−\mathrm{1}\right)^{\mathrm{2}} }}{\mathrm{3}} \\ $$$${x}=\frac{\mathrm{2}{a}+\mathrm{2}\mid{a}−\mathrm{1}\mid}{\mathrm{3}}=\frac{\mathrm{2}{a}−\mathrm{2}\left({a}−\mathrm{1}\right)}{\mathrm{3}}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\leftarrow{the}\:{answer} \\ $$$${note}\:\mid{a}−\mathrm{1}\mid\:=\:−\left({a}−\mathrm{1}\right)\:{for}\:\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by ZiYangLee last updated on 25/Aug/20
Why log_a (4a(1−2a)) at the second row?
$$\mathrm{Why}\:\mathrm{log}_{{a}} \left(\mathrm{4}{a}\left(\mathrm{1}−\mathrm{2}{a}\right)\right)\:\mathrm{at}\:\mathrm{the}\:\mathrm{second}\:\mathrm{row}? \\ $$$$ \\ $$
Commented by bemath last updated on 25/Aug/20
(1/(log _2 (a))) = log _a (2)
$$\frac{\mathrm{1}}{\mathrm{log}\:_{\mathrm{2}} \left({a}\right)}\:=\:\mathrm{log}\:_{{a}} \left(\mathrm{2}\right) \\ $$
Commented by Coronavirus last updated on 25/Aug/20
log_2 (a)=((ln(a) )/(ln(2) ))⇒(2/(log_2 (a)))=((2ln(2) )/(ln(a) ))=((ln(4) )/(ln(a) ))=log_a (4)
$$\mathrm{log}_{\mathrm{2}} \left(\mathrm{a}\right)=\frac{\mathrm{ln}\left(\mathrm{a}\right)\:}{\mathrm{ln}\left(\mathrm{2}\right)\:}\Rightarrow\frac{\mathrm{2}}{\mathrm{log}_{\mathrm{2}} \left(\mathrm{a}\right)}=\frac{\mathrm{2ln}\left(\mathrm{2}\right)\:}{\mathrm{ln}\left(\mathrm{a}\right)\:}=\frac{\mathrm{ln}\left(\mathrm{4}\right)\:}{\mathrm{ln}\left(\mathrm{a}\right)\:}=\mathrm{log}_{\mathrm{a}} \left(\mathrm{4}\right) \\ $$
Answered by Rasheed.Sindhi last updated on 25/Aug/20
log_a (3x−4a)+log_a 3x−log_a (1−2a) =(2/(log_2 a)) log_a (((3x(3x−4a))/((1−2a))))=(2/( (1/(log_a 2)) )) log_a (((3x(3x−4a))/((1−2a))))=2log_a 2 log_a (((3x(3x−4a))/((1−2a))))=log_a 2^2 ((3x(3x−4a))/((1−2a)))=2^2 9x^2 −12ax=4(1−2a) 9x^2 −12ax−4(1−2a)=0 x=((12a±(√(144a^2 −4(9)(−4(1−2a))))/(2(9))) x=((12a±(√(144a^2 +144(1−2a))))/(18)) x=((12a±12(√(a^2 −2a+1)))/(18)) x=((2a±2(a−1))/3)=((2a±2a∓2)/3) x=4a−2 , 2 ◂
$$\mathrm{log}_{{a}} \left(\mathrm{3}{x}−\mathrm{4}{a}\right)+\mathrm{log}_{{a}} \mathrm{3}{x}−\mathrm{log}_{{a}} \left(\mathrm{1}−\mathrm{2}{a}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}}{\mathrm{log}_{\mathrm{2}} {a}} \\ $$$$\mathrm{log}_{{a}} \left(\frac{\mathrm{3}{x}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)}{\left(\mathrm{1}−\mathrm{2}{a}\right)}\right)=\frac{\mathrm{2}}{\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{log}_{{a}} \mathrm{2}}\:\:\:\:} \\ $$$$\mathrm{log}_{{a}} \left(\frac{\mathrm{3}{x}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)}{\left(\mathrm{1}−\mathrm{2}{a}\right)}\right)=\mathrm{2log}_{{a}} \mathrm{2} \\ $$$$\mathrm{log}_{{a}} \left(\frac{\mathrm{3}{x}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)}{\left(\mathrm{1}−\mathrm{2}{a}\right)}\right)=\mathrm{log}_{{a}} \mathrm{2}^{\mathrm{2}} \\ $$$$\frac{\mathrm{3}{x}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)}{\left(\mathrm{1}−\mathrm{2}{a}\right)}=\mathrm{2}^{\mathrm{2}} \\ $$$$\mathrm{9}{x}^{\mathrm{2}} −\mathrm{12}{ax}=\mathrm{4}\left(\mathrm{1}−\mathrm{2}{a}\right) \\ $$$$\mathrm{9}{x}^{\mathrm{2}} −\mathrm{12}{ax}−\mathrm{4}\left(\mathrm{1}−\mathrm{2}{a}\right)=\mathrm{0} \\ $$$${x}=\frac{\mathrm{12}{a}\pm\sqrt{\mathrm{144}{a}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{9}\right)\left(−\mathrm{4}\left(\mathrm{1}−\mathrm{2}{a}\right)\right.}}{\mathrm{2}\left(\mathrm{9}\right)} \\ $$$${x}=\frac{\mathrm{12}{a}\pm\sqrt{\mathrm{144}{a}^{\mathrm{2}} +\mathrm{144}\left(\mathrm{1}−\mathrm{2}{a}\right)}}{\mathrm{18}} \\ $$$${x}=\frac{\mathrm{12}{a}\pm\mathrm{12}\sqrt{{a}^{\mathrm{2}} −\mathrm{2}{a}+\mathrm{1}}}{\mathrm{18}} \\ $$$${x}=\frac{\mathrm{2}{a}\pm\mathrm{2}\left({a}−\mathrm{1}\right)}{\mathrm{3}}=\frac{\mathrm{2}{a}\pm\mathrm{2}{a}\mp\mathrm{2}}{\mathrm{3}} \\ $$$${x}=\mathrm{4}{a}−\mathrm{2}\:,\:\mathrm{2}\:\:\:\blacktriangleleft\:\: \\ $$$$ \\ $$
Answered by Her_Majesty last updated on 25/Aug/20
a>0∧a≠1∧1−2a>0∧1−2a≠1 ⇔ 0<a<(1/2) 3x>0∧3x−4a>0 ⇔ x>((4a)/3) ((ln(3x−4a))/(lna))+((ln3x)/(lna))=((2ln2)/(lna))+((ln(1−2a))/(lna)) ln(3x(3x−4a))=ln(4(1−2a)) 3x(3x−4a)=4(1−2a) x^2 −((4a)/3)x+((4(2a−1))/9)=0 (x−(2/3))(x−((2(2a−1))/3))=0 x_1 =(2/3) x_2 =((2(2a−1))/3) but 0<a<(1/2)∧x>((4a)/3) makes this impossible: x>((4a)/3)∧x=((4a)/3)−(1/3) is false
$${a}>\mathrm{0}\wedge{a}\neq\mathrm{1}\wedge\mathrm{1}−\mathrm{2}{a}>\mathrm{0}\wedge\mathrm{1}−\mathrm{2}{a}\neq\mathrm{1}\:\Leftrightarrow\:\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{3}{x}>\mathrm{0}\wedge\mathrm{3}{x}−\mathrm{4}{a}>\mathrm{0}\:\Leftrightarrow\:{x}>\frac{\mathrm{4}{a}}{\mathrm{3}} \\ $$$$\frac{{ln}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)}{{lna}}+\frac{{ln}\mathrm{3}{x}}{{lna}}=\frac{\mathrm{2}{ln}\mathrm{2}}{{lna}}+\frac{{ln}\left(\mathrm{1}−\mathrm{2}{a}\right)}{{lna}} \\ $$$${ln}\left(\mathrm{3}{x}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)\right)={ln}\left(\mathrm{4}\left(\mathrm{1}−\mathrm{2}{a}\right)\right) \\ $$$$\mathrm{3}{x}\left(\mathrm{3}{x}−\mathrm{4}{a}\right)=\mathrm{4}\left(\mathrm{1}−\mathrm{2}{a}\right) \\ $$$${x}^{\mathrm{2}} −\frac{\mathrm{4}{a}}{\mathrm{3}}{x}+\frac{\mathrm{4}\left(\mathrm{2}{a}−\mathrm{1}\right)}{\mathrm{9}}=\mathrm{0} \\ $$$$\left({x}−\frac{\mathrm{2}}{\mathrm{3}}\right)\left({x}−\frac{\mathrm{2}\left(\mathrm{2}{a}−\mathrm{1}\right)}{\mathrm{3}}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}_{\mathrm{2}} =\frac{\mathrm{2}\left(\mathrm{2}{a}−\mathrm{1}\right)}{\mathrm{3}}\:{but}\:\mathrm{0}<{a}<\frac{\mathrm{1}}{\mathrm{2}}\wedge{x}>\frac{\mathrm{4}{a}}{\mathrm{3}}\:{makes} \\ $$$${this}\:{impossible}: \\ $$$${x}>\frac{\mathrm{4}{a}}{\mathrm{3}}\wedge{x}=\frac{\mathrm{4}{a}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{3}}\:{is}\:{false} \\ $$

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