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Question Number 17386 by ajfour last updated on 05/Jul/17

Solve the equation: log _2 x log _3 x log _5 x=log _2 x log _3 x +log _3 x log _5 x +log _5 x log _2 x .

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{3}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x}=\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{3}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{log}\:_{\mathrm{3}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{5}} \mathrm{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{log}\:_{\mathrm{5}} \mathrm{x}\:\mathrm{log}\:_{\mathrm{2}} \mathrm{x}\:. \\ $$

Answered by mrW1 last updated on 05/Jul/17

let u=ln x a=ln 2 b=ln 3 c=ln 5 ⇒(u^3 /(abc))=(u^2 /(ab))+(u^2 /(bc))+(u^2 /(ca)) ⇒u^3 =(a+b+c)u^2 ⇒u^2 [u−(a+b+c)]=0 ⇒u=0 or ⇒u=a+b+c ⇒ln x=0 ⇒x=1 ⇒ln x=ln 2+ln 3+ln 5=ln (2×3×5) ⇒x=2×3×5=30

$$\mathrm{let}\:\mathrm{u}=\mathrm{ln}\:\mathrm{x} \\ $$$$\mathrm{a}=\mathrm{ln}\:\mathrm{2} \\ $$$$\mathrm{b}=\mathrm{ln}\:\mathrm{3} \\ $$$$\mathrm{c}=\mathrm{ln}\:\mathrm{5} \\ $$$$\Rightarrow\frac{\mathrm{u}^{\mathrm{3}} }{\mathrm{abc}}=\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{ab}}+\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{bc}}+\frac{\mathrm{u}^{\mathrm{2}} }{\mathrm{ca}} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{3}} =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\mathrm{u}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{2}} \left[\mathrm{u}−\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\right]=\mathrm{0} \\ $$$$\Rightarrow\mathrm{u}=\mathrm{0} \\ $$$$\mathrm{or} \\ $$$$\Rightarrow\mathrm{u}=\mathrm{a}+\mathrm{b}+\mathrm{c} \\ $$$$ \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{x}=\mathrm{1} \\ $$$$ \\ $$$$\Rightarrow\mathrm{ln}\:\mathrm{x}=\mathrm{ln}\:\mathrm{2}+\mathrm{ln}\:\mathrm{3}+\mathrm{ln}\:\mathrm{5}=\mathrm{ln}\:\left(\mathrm{2}×\mathrm{3}×\mathrm{5}\right) \\ $$$$\Rightarrow\mathrm{x}=\mathrm{2}×\mathrm{3}×\mathrm{5}=\mathrm{30} \\ $$

Commented by ajfour last updated on 05/Jul/17

great, you play confident sir! who else can ?

$$\mathrm{great},\:\mathrm{you}\:\mathrm{play}\:\mathrm{confident}\:\mathrm{sir}! \\ $$$$\mathrm{who}\:\mathrm{else}\:\mathrm{can}\:? \\ $$

Commented by 47358857 last updated on 07/Dec/17

Answered by b.e.h.i.8.3.4.17@gmail.com last updated on 06/Jul/17

abc=ab+bc+ca ⇒(1/a)+(1/b)+(1/c)=1⇒log_x ^2 +log_x ^3 +log_x ^5 =1 ⇒log_x ^(2×3×5) =1⇒log_x ^(30) =1⇒x=30 . also: x=1,can be an answer.

$${abc}={ab}+{bc}+{ca} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\mathrm{1}\Rightarrow{log}_{{x}} ^{\mathrm{2}} +{log}_{{x}} ^{\mathrm{3}} +{log}_{{x}} ^{\mathrm{5}} =\mathrm{1} \\ $$$$\Rightarrow{log}_{{x}} ^{\mathrm{2}×\mathrm{3}×\mathrm{5}} \:=\mathrm{1}\Rightarrow{log}_{{x}} ^{\mathrm{30}} =\mathrm{1}\Rightarrow{x}=\mathrm{30}\:. \\ $$$${also}:\:{x}=\mathrm{1},{can}\:{be}\:{an}\:{answer}. \\ $$

Commented by ajfour last updated on 08/Jul/17

very nice way, sir.

$$\mathrm{very}\:\mathrm{nice}\:\mathrm{way},\:\mathrm{sir}. \\ $$

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