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Question Number 55224 by Otchere Abdullai last updated on 19/Feb/19
log_2 (x^2 +7x−2)=log_2 (x^2 +3x−6)+log_4 8 find x
$${log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}\right)={log}_{\mathrm{2}} \left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}\right)+{log}_{\mathrm{4}} \mathrm{8} \\ $$$${find}\:{x} \\ $$
Answered by peter frank last updated on 19/Feb/19
log_2 (((x^2 +7x−2)/(x^2 +3x−6)))=3log_2^2 2 log_2 (((x^2 +7x−2)/(x^2 +3x−6)))=((log 8)/(log 4)) log_2 (((x^2 +7x−2)/(x^2 +3x−6)))=((3log 2)/(2log 2)) log_2 (((x^2 +7x−2)/(x^2 +3x−6)))=(3/2) 2^(3/2) =((x^2 +7x−2)/(x^2 +3x−6)) ........
$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{6}}\right)=\mathrm{3log}_{\mathrm{2}^{\mathrm{2}} } \mathrm{2} \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{6}}\right)=\frac{\mathrm{log}\:\mathrm{8}}{\mathrm{log}\:\mathrm{4}} \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{6}}\right)=\frac{\mathrm{3log}\:\mathrm{2}}{\mathrm{2log}\:\mathrm{2}} \\ $$$$\mathrm{log}_{\mathrm{2}} \left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{6}}\right)=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{2}^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{7x}−\mathrm{2}}{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{6}} \\ $$$$…….. \\ $$$$ \\ $$
Commented by Otchere Abdullai last updated on 19/Feb/19
thanks sir but in the book the answer is 7 but was not solved
$${thanks}\:{sir}\:{but}\:{in}\:{the}\:{book}\:{the}\:{answer} \\ $$$${is}\:\mathrm{7}\:{but}\:{was}\:{not}\:{solved} \\ $$
Answered by kaivan.ahmadi last updated on 19/Feb/19
log_2 (((x^2 +7x−2)/(x^2 +3x−6)))=log_2^2 8=(1/2)log_2 8=log_2 (√8)⇒ x^2 +7x−2=(√8)x^2 +3(√8)x−6(√8)⇒ ((√8)−1)x^2 +(3(√8)−7)x−(6(√8)−2)=0 Δ=(3(√8)−7)^2 −4((√8)−1)(6(√8)−2)= 72−42(√8)+49−4(48−8(√8)+2)= 121−42(√8)−192+32(√8)−8= −79−10(√8)<0 ⇒there is no real number for x
$$ \\ $$$${log}_{\mathrm{2}} \left(\frac{{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}}\right)={log}_{\mathrm{2}^{\mathrm{2}} } \mathrm{8}=\frac{\mathrm{1}}{\mathrm{2}}{log}_{\mathrm{2}} \mathrm{8}={log}_{\mathrm{2}} \sqrt{\mathrm{8}}\Rightarrow \\ $$$${x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}=\sqrt{\mathrm{8}}{x}^{\mathrm{2}} +\mathrm{3}\sqrt{\mathrm{8}}{x}−\mathrm{6}\sqrt{\mathrm{8}}\Rightarrow \\ $$$$\left(\sqrt{\mathrm{8}}−\mathrm{1}\right){x}^{\mathrm{2}} +\left(\mathrm{3}\sqrt{\mathrm{8}}−\mathrm{7}\right){x}−\left(\mathrm{6}\sqrt{\mathrm{8}}−\mathrm{2}\right)=\mathrm{0} \\ $$$$\Delta=\left(\mathrm{3}\sqrt{\mathrm{8}}−\mathrm{7}\right)^{\mathrm{2}} −\mathrm{4}\left(\sqrt{\mathrm{8}}−\mathrm{1}\right)\left(\mathrm{6}\sqrt{\mathrm{8}}−\mathrm{2}\right)= \\ $$$$\mathrm{72}−\mathrm{42}\sqrt{\mathrm{8}}+\mathrm{49}−\mathrm{4}\left(\mathrm{48}−\mathrm{8}\sqrt{\mathrm{8}}+\mathrm{2}\right)= \\ $$$$\mathrm{121}−\mathrm{42}\sqrt{\mathrm{8}}−\mathrm{192}+\mathrm{32}\sqrt{\mathrm{8}}−\mathrm{8}= \\ $$$$−\mathrm{79}−\mathrm{10}\sqrt{\mathrm{8}}<\mathrm{0} \\ $$$$\Rightarrow{there}\:{is}\:{no}\:{real}\:{number}\:{for}\:{x} \\ $$$$ \\ $$
Commented by Otchere Abdullai last updated on 19/Feb/19
thanks sir the answer in the book is x=7 but unsolved question
$${thanks}\:{sir}\:{the}\:{answer}\:{in}\:{the}\:{book}\:{is}\: \\ $$$${x}=\mathrm{7}\:{but}\:{unsolved}\:{question} \\ $$
Commented by kaivan.ahmadi last updated on 19/Feb/19
but if you replace x=7 the equality is not true
$${but}\:{if}\:{you}\:{replace}\:{x}=\mathrm{7}\:{the}\:{equality}\:{is}\:{not}\:{true} \\ $$
Commented by Otchere Abdullai last updated on 19/Feb/19
ok thanks sir i will re−check the question from the library
$${ok}\:{thanks}\:{sir}\:{i}\:{will}\:{re}−{check}\:{the} \\ $$$${question}\:{from}\:{the}\:{library} \\ $$
Answered by MJS last updated on 20/Feb/19
log_2 a =((ln a)/(ln 2)) log_4 8 =((ln 8)/(ln 4))=((3ln 2)/(2ln 2))=(3/2) ln (x^2 +7x−2)=ln (x^2 +3x−6)+((3ln 2)/2) ln ((x^2 +7x−2)/(x^2 +3x−6)) =((3ln 2)/2) ((x^2 +7x−2)/(x^2 +3x−6))=2(√2) transforming x^2 +((17−8(√2))/7)x−((2(23+4(√2)))/7)=0 x=−((17−8(√2))/(14))±((√(1705−48(√2)))/(14)) both satisfy the given equation although the “−” solution leads to complex values for both logarithms
$$\mathrm{log}_{\mathrm{2}} \:{a}\:=\frac{\mathrm{ln}\:{a}}{\mathrm{ln}\:\mathrm{2}} \\ $$$$\mathrm{log}_{\mathrm{4}} \:\mathrm{8}\:=\frac{\mathrm{ln}\:\mathrm{8}}{\mathrm{ln}\:\mathrm{4}}=\frac{\mathrm{3ln}\:\mathrm{2}}{\mathrm{2ln}\:\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}\right)=\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}\right)+\frac{\mathrm{3ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:\frac{{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}}\:=\frac{\mathrm{3ln}\:\mathrm{2}}{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{7}{x}−\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{3}{x}−\mathrm{6}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\mathrm{transforming} \\ $$$${x}^{\mathrm{2}} +\frac{\mathrm{17}−\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{7}}{x}−\frac{\mathrm{2}\left(\mathrm{23}+\mathrm{4}\sqrt{\mathrm{2}}\right)}{\mathrm{7}}=\mathrm{0} \\ $$$${x}=−\frac{\mathrm{17}−\mathrm{8}\sqrt{\mathrm{2}}}{\mathrm{14}}\pm\frac{\sqrt{\mathrm{1705}−\mathrm{48}\sqrt{\mathrm{2}}}}{\mathrm{14}} \\ $$$$\mathrm{both}\:\mathrm{satisfy}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation}\:\mathrm{although} \\ $$$$\mathrm{the}\:“−''\:\mathrm{solution}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{complex}\:\mathrm{values} \\ $$$$\mathrm{for}\:\mathrm{both}\:\mathrm{logarithms} \\ $$
Commented by Otchere Abdullai last updated on 20/Feb/19
Thank you mjs sir!
$${Thank}\:{you}\:{mjs}\:{sir}! \\ $$

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