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Question Number 221853 by fantastic last updated on 11/Jun/25

find x where log _8 x−log _4 x−log _2 x=11

$${find}\:{x}\:{where} \\ $$$$\mathrm{log}\underset{\mathrm{8}} {\:}{x}−\mathrm{log}\underset{\mathrm{4}} {\:}{x}−\mathrm{log}\underset{\mathrm{2}} {\:}{x}=\mathrm{11} \\ $$

Commented by Tawa11 last updated on 11/Jun/25

log_2^3 x − log_2^2 x − log_2 x = 11 ∴ (1/3)log_2 x − (1/2)log_2 x − log_2 x = 11 ∴ − (7/6)log_2 x = 11 ∴ log_2 x^(− (7/6)) = 11 ∴ x^(− (7/6)) = 2^(11) ∴ x^(− (7/6) × − (6/7)) = 2^(11 × − (6/7)) ∴ x = 2^(− ((66)/7))

$$\mathrm{log}_{\mathrm{2}^{\mathrm{3}} } \mathrm{x}\:\:−\:\:\mathrm{log}_{\mathrm{2}^{\mathrm{2}} } \mathrm{x}\:\:\:−\:\:\:\mathrm{log}_{\mathrm{2}} \mathrm{x}\:\:\:=\:\:\mathrm{11} \\ $$$$\therefore\:\:\frac{\mathrm{1}}{\mathrm{3}}\mathrm{log}_{\mathrm{2}} \mathrm{x}\:\:−\:\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}_{\mathrm{2}} \mathrm{x}\:\:−\:\:\mathrm{log}_{\mathrm{2}} \mathrm{x}\:\:=\:\:\mathrm{11} \\ $$$$\therefore\:\:\:\:\:\:−\:\frac{\mathrm{7}}{\mathrm{6}}\mathrm{log}_{\mathrm{2}} \mathrm{x}\:\:\:=\:\:\mathrm{11} \\ $$$$\therefore\:\:\:\:\:\:\:\:\mathrm{log}_{\mathrm{2}} \mathrm{x}^{−\:\frac{\mathrm{7}}{\mathrm{6}}} \:\:\:=\:\:\mathrm{11} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{−\:\frac{\mathrm{7}}{\mathrm{6}}} \:\:\:=\:\:\mathrm{2}^{\mathrm{11}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\mathrm{x}^{−\:\frac{\mathrm{7}}{\mathrm{6}}\:×\:−\:\frac{\mathrm{6}}{\mathrm{7}}} \:\:\:=\:\:\mathrm{2}^{\mathrm{11}\:×\:−\:\frac{\mathrm{6}}{\mathrm{7}}} \\ $$$$\therefore\:\:\:\:\:\:\:\:\:\:\:\mathrm{x}\:\:\:=\:\:\mathrm{2}^{−\:\frac{\mathrm{66}}{\mathrm{7}}} \\ $$

Commented by fantastic last updated on 11/Jun/25

thanks sir

$${thanks}\:{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 13/Jun/25

Miss Tawa, please send your full solution as ′answer′ not as ′comment′.

$${Miss}\:{Tawa},\:{please}\:{send}\:{your}\: \\ $$$${full}\:{solution}\:{as}\:'{answer}'\:{not} \\ $$$${as}\:'{comment}'. \\ $$

Answered by Rasheed.Sindhi last updated on 11/Jun/25

((log_2 x )/(log_2 2^3 ))−((log_2 x)/(log_2 2^2 ))−((log_2 x)/(log_2 2))=11 ((log_2 x )/3)−((log_2 x)/2)−((log_2 x)/1)=11 2log_2 x−3log_2 x−6log_2 x=66 log_2 x(2−3−6)=66 log_2 x=−((66)/7) x=2^(−((66)/7))

$$\frac{\mathrm{log}_{\mathrm{2}} {x}\:}{\mathrm{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{3}} }−\frac{\mathrm{log}_{\mathrm{2}} {x}}{\mathrm{log}_{\mathrm{2}} \mathrm{2}^{\mathrm{2}} }−\frac{\mathrm{log}_{\mathrm{2}} {x}}{\mathrm{log}_{\mathrm{2}} \mathrm{2}}=\mathrm{11} \\ $$$$\frac{\mathrm{log}_{\mathrm{2}} {x}\:}{\mathrm{3}}−\frac{\mathrm{log}_{\mathrm{2}} {x}}{\mathrm{2}}−\frac{\mathrm{log}_{\mathrm{2}} {x}}{\mathrm{1}}=\mathrm{11} \\ $$$$\mathrm{2log}_{\mathrm{2}} {x}−\mathrm{3log}_{\mathrm{2}} {x}−\mathrm{6log}_{\mathrm{2}} {x}=\mathrm{66} \\ $$$$\mathrm{log}_{\mathrm{2}} {x}\left(\mathrm{2}−\mathrm{3}−\mathrm{6}\right)=\mathrm{66} \\ $$$$\mathrm{log}_{\mathrm{2}} {x}=−\frac{\mathrm{66}}{\mathrm{7}} \\ $$$${x}=\mathrm{2}^{−\frac{\mathrm{66}}{\mathrm{7}}} \\ $$

Commented by fantastic last updated on 11/Jun/25

thanks sir

$${thanks}\:{sir} \\ $$

Answered by shunmisaki007 last updated on 11/Jun/25

log_8 (x)−log_4 (x)−log_2 (x)=11 ((ln(x))/(ln(8)))−((ln(x))/(ln(4)))−((ln(x))/(ln(2)))=11 ((ln(x))/(3ln(2)))−((ln(x))/(2ln(2)))−((ln(x))/(ln(2)))=11 ((2ln(x)−3ln(x)−6ln(x))/(6ln(2)))=11 −7ln(x)=66ln(2) ln(x)=−((66)/7)ln(2) x=2^(−((66)/7)) =(1/2^((66)/7) ) Ans.

$$\mathrm{log}_{\mathrm{8}} \left({x}\right)−\mathrm{log}_{\mathrm{4}} \left({x}\right)−\mathrm{log}_{\mathrm{2}} \left({x}\right)=\mathrm{11} \\ $$$$\frac{\mathrm{ln}\left({x}\right)}{\mathrm{ln}\left(\mathrm{8}\right)}−\frac{\mathrm{ln}\left({x}\right)}{\mathrm{ln}\left(\mathrm{4}\right)}−\frac{\mathrm{ln}\left({x}\right)}{\mathrm{ln}\left(\mathrm{2}\right)}=\mathrm{11} \\ $$$$\frac{\mathrm{ln}\left({x}\right)}{\mathrm{3ln}\left(\mathrm{2}\right)}−\frac{\mathrm{ln}\left({x}\right)}{\mathrm{2ln}\left(\mathrm{2}\right)}−\frac{\mathrm{ln}\left({x}\right)}{\mathrm{ln}\left(\mathrm{2}\right)}=\mathrm{11} \\ $$$$\frac{\mathrm{2ln}\left({x}\right)−\mathrm{3ln}\left({x}\right)−\mathrm{6ln}\left({x}\right)}{\mathrm{6ln}\left(\mathrm{2}\right)}=\mathrm{11} \\ $$$$−\mathrm{7ln}\left({x}\right)=\mathrm{66ln}\left(\mathrm{2}\right) \\ $$$$\mathrm{ln}\left({x}\right)=−\frac{\mathrm{66}}{\mathrm{7}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$${x}=\mathrm{2}^{−\frac{\mathrm{66}}{\mathrm{7}}} =\frac{\mathrm{1}}{\mathrm{2}^{\frac{\mathrm{66}}{\mathrm{7}}} }\:\boldsymbol{{Ans}}. \\ $$

Commented by fantastic last updated on 11/Jun/25

thanks sir

$${thanks}\:{sir} \\ $$

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