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a-3-2-b-1-5-1-6-6-and-x-y-R-such-that-3x-2y-log-a-18-5-4-2x-y-log-b-1080-then-find-the-value-of-4x-5y-

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Question Number 222142 by fantastic last updated on 19/Jun/25
a=3(√2) ,b=(1/(5^(1/6) (√6))) and x,yεR such that 3x +2y=log _a (18)^(5/4) 2x−y=log _b ((√(1080))) then find the value of 4x+5y
$${a}=\mathrm{3}\sqrt{\mathrm{2}}\:,{b}=\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{6}}} \sqrt{\mathrm{6}}}\:{and}\:{x},{y}\epsilon\mathbb{R}\:{such}\:{that} \\ $$$$\mathrm{3}{x}\:+\mathrm{2}{y}=\mathrm{log}\:_{{a}} \left(\mathrm{18}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\mathrm{2}{x}−{y}=\mathrm{log}\:_{{b}} \left(\sqrt{\mathrm{1080}}\right) \\ $$$${then}\:{find}\:{the}\:{value}\:{of}\:\: \\ $$$$\mathrm{4}{x}+\mathrm{5}{y} \\ $$
Answered by Rasheed.Sindhi last updated on 19/Jun/25
3x +2y=log _a (18)^(5/4) 3x+2y=(5/4)log_a 18 12x+8y=5log_(3(√2)) (2×3^2 ) ((12x+8y)/5) =((log 2+2log 3 )/(log 3+(1/2)log 2)) ((12x+8y)/5) =((log 2+2log 3 )/((1/2)(2log 3+log 2)))=2 determinant (((6x+4y=5)))....(i) 2x−y=log _b ((√(1080))) 2x−y=(1/2)log _b 1080 log _b 1080=((log 1080)/(log (1/(5^(1/6) (√6))))) =((log(2^3 .3^3 .5) )/(log(5^(− (1/6)) .6^(− (1/2)) ))) =((3log 2+3log 3+log 5)/(−(1/6)log 5−(1/2)log 6)) =−((3log 2+3log 3+log 5)/((1/6)log 5+(1/2)log(2.3))) =−((3log 2+3log 3+log 5)/((1/6)log 5+(1/2)log 2+(1/2)log 3))) =−((3log 2+3log 3+log 5)/((1/6)(log 5+3log 2+3log 3)))=−6 2x−y=(1/2)log _b 1080=(1/2)(−6) determinant (((2x−y=−3))).....(ii) (i) & (ii): { ((6x+4y=5)),((2x−y=−3)) :} ⇒y=((14)/7)=2 ,x=−(1/2) 4x+5y=4(−(1/2))+5(2) =−2+10=8
$$\mathrm{3}{x}\:+\mathrm{2}{y}=\mathrm{log}\:_{{a}} \left(\mathrm{18}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} \\ $$$$\mathrm{3}{x}+\mathrm{2}{y}=\frac{\mathrm{5}}{\mathrm{4}}\mathrm{log}_{{a}} \mathrm{18} \\ $$$$\mathrm{12}{x}+\mathrm{8}{y}=\mathrm{5log}_{\mathrm{3}\sqrt{\mathrm{2}}} \left(\mathrm{2}×\mathrm{3}^{\mathrm{2}} \right) \\ $$$$\frac{\mathrm{12}{x}+\mathrm{8}{y}}{\mathrm{5}}\:=\frac{\mathrm{log}\:\mathrm{2}+\mathrm{2log}\:\mathrm{3}\:}{\mathrm{log}\:\mathrm{3}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mathrm{2}} \\ $$$$\frac{\mathrm{12}{x}+\mathrm{8}{y}}{\mathrm{5}}\:=\frac{\mathrm{log}\:\mathrm{2}+\mathrm{2log}\:\mathrm{3}\:}{\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2log}\:\mathrm{3}+\mathrm{log}\:\mathrm{2}\right)}=\mathrm{2} \\ $$$$\:\begin{array}{|c|}{\mathrm{6}{x}+\mathrm{4}{y}=\mathrm{5}}\\\hline\end{array}….\left({i}\right) \\ $$$$\: \\ $$$$\mathrm{2}{x}−{y}=\mathrm{log}\:_{{b}} \left(\sqrt{\mathrm{1080}}\right) \\ $$$$\mathrm{2}{x}−{y}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:_{{b}} \mathrm{1080} \\ $$$$\mathrm{log}\:_{{b}} \mathrm{1080}=\frac{\mathrm{log}\:\mathrm{1080}}{\mathrm{log}\:\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{6}}} \sqrt{\mathrm{6}}}} \\ $$$$\:\:\:=\frac{\mathrm{log}\left(\mathrm{2}^{\mathrm{3}} .\mathrm{3}^{\mathrm{3}} .\mathrm{5}\right)\:}{\mathrm{log}\left(\mathrm{5}^{−\:\frac{\mathrm{1}}{\mathrm{6}}} .\mathrm{6}^{−\:\frac{\mathrm{1}}{\mathrm{2}}} \:\right)} \\ $$$$\:\:=\frac{\mathrm{3log}\:\mathrm{2}+\mathrm{3log}\:\mathrm{3}+\mathrm{log}\:\mathrm{5}}{−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{log}\:\mathrm{5}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mathrm{6}} \\ $$$$\:\:=−\frac{\mathrm{3log}\:\mathrm{2}+\mathrm{3log}\:\mathrm{3}+\mathrm{log}\:\mathrm{5}}{\frac{\mathrm{1}}{\mathrm{6}}\mathrm{log}\:\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\left(\mathrm{2}.\mathrm{3}\right)} \\ $$$$\:\:=−\frac{\mathrm{3log}\:\mathrm{2}+\mathrm{3log}\:\mathrm{3}+\mathrm{log}\:\mathrm{5}}{\left.\frac{\mathrm{1}}{\mathrm{6}}\mathrm{log}\:\mathrm{5}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mathrm{2}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:\mathrm{3}\right)} \\ $$$$\:\:=−\frac{\cancel{\mathrm{3log}\:\mathrm{2}+\mathrm{3log}\:\mathrm{3}+\mathrm{log}\:\mathrm{5}}}{\frac{\mathrm{1}}{\mathrm{6}}\left(\cancel{\mathrm{log}\:\mathrm{5}+\mathrm{3log}\:\mathrm{2}+\mathrm{3log}\:\mathrm{3}}\right)}=−\mathrm{6} \\ $$$$\mathrm{2}{x}−{y}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:_{{b}} \mathrm{1080}=\frac{\mathrm{1}}{\mathrm{2}}\left(−\mathrm{6}\right) \\ $$$$\:\begin{array}{|c|}{\mathrm{2}{x}−{y}=−\mathrm{3}}\\\hline\end{array}…..\left({ii}\right) \\ $$$$\left({i}\right)\:\&\:\left({ii}\right): \\ $$$$\begin{cases}{\mathrm{6}{x}+\mathrm{4}{y}=\mathrm{5}}\\{\mathrm{2}{x}−{y}=−\mathrm{3}}\end{cases}\:\Rightarrow{y}=\frac{\mathrm{14}}{\mathrm{7}}=\mathrm{2}\:,{x}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{4}{x}+\mathrm{5}{y}=\mathrm{4}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{5}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\mathrm{2}+\mathrm{10}=\mathrm{8} \\ $$
Answered by som(math1967) last updated on 19/Jun/25
a=3(√2)=(√(18)) log_(√(18)) (18)^(5/4) =(5/2)log_(√(18)) (√(18))=(5/2) ∴3x+2y=(5/2)⇒6x+4y=5 ...case1 b=(1/(5^(1/6) (√6)))=(1/((5×216)^(1/6) ))=(1080)^(−(1/6)) log_((1080)^(−(1/6)) ) (√(1080)) =log_((1080)^(−(1/6)) ) (1080)^((−3)/(−6)) =−3 ∴2x−y=−3 ....case2 case1 −case2 4x+5y=5+3=8
$$\:{a}=\mathrm{3}\sqrt{\mathrm{2}}=\sqrt{\mathrm{18}} \\ $$$$\:{log}_{\sqrt{\mathrm{18}}} \left(\mathrm{18}\right)^{\frac{\mathrm{5}}{\mathrm{4}}} =\frac{\mathrm{5}}{\mathrm{2}}{log}_{\sqrt{\mathrm{18}}} \sqrt{\mathrm{18}}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\therefore\mathrm{3}{x}+\mathrm{2}{y}=\frac{\mathrm{5}}{\mathrm{2}}\Rightarrow\mathrm{6}{x}+\mathrm{4}{y}=\mathrm{5}\:\:\:…{case}\mathrm{1} \\ $$$${b}=\frac{\mathrm{1}}{\mathrm{5}^{\frac{\mathrm{1}}{\mathrm{6}}} \sqrt{\mathrm{6}}}=\frac{\mathrm{1}}{\left(\mathrm{5}×\mathrm{216}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} }=\left(\mathrm{1080}\right)^{−\frac{\mathrm{1}}{\mathrm{6}}} \\ $$$${log}_{\left(\mathrm{1080}\right)^{−\frac{\mathrm{1}}{\mathrm{6}}} } \sqrt{\mathrm{1080}} \\ $$$$={log}_{\left(\mathrm{1080}\right)^{−\frac{\mathrm{1}}{\mathrm{6}}} } \left(\mathrm{1080}\right)^{\frac{−\mathrm{3}}{−\mathrm{6}}} =−\mathrm{3} \\ $$$$\therefore\mathrm{2}{x}−{y}=−\mathrm{3}\:\:….{case}\mathrm{2} \\ $$$${case}\mathrm{1}\:−{case}\mathrm{2} \\ $$$$\mathrm{4}{x}+\mathrm{5}{y}=\mathrm{5}+\mathrm{3}=\mathrm{8} \\ $$$$ \\ $$$$ \\ $$
Commented by fantastic last updated on 19/Jun/25
nice!
$${nice}! \\ $$$$ \\ $$
Commented by AgniMath last updated on 22/Jun/25
JEE ADV 2024 QS
$${JEE}\:{ADV}\:\:\mathrm{2024}\:{QS} \\ $$

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