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If-x-log-a-bc-y-log-b-ca-z-log-c-ab-prove-that-x-y-z-xyz-2-

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Question Number 223513 by fantastic last updated on 27/Jul/25
If x=log _a bc , y=log _b ca , z=log _c ab prove that x+y+z=xyz−2
$${If}\:{x}=\mathrm{log}\:_{{a}} {bc}\:,\:{y}=\mathrm{log}\:_{{b}} {ca}\:,\:{z}=\mathrm{log}\:_{{c}} {ab} \\ $$$${prove}\:{that}\:{x}+{y}+{z}={xyz}−\mathrm{2} \\ $$
Answered by som(math1967) last updated on 27/Jul/25
1+x=1+log_a bc=log_a a+log_a bc 1+x=log_a abc⇒(1/(1+x))=log_(abc) a (1/(1+y))=log_(abc) b, (1/(1+y))=log_(abc) c ∴ (1/(1+x))+(1/(1+y)) +(1/(1+z))=log_(abc) abc=1 (1/(1+x)) +(1/(1+y))=1−(1/(1+z)) ⇒((1+y+1+x)/(1+y+x+xy))=((1+z−1)/(1+z)) ⇒(2+x+y)(1+z)=z(1+x+y+xy) ⇒2+2z+x+xz+y+yz =z+zx+yz+xyz ⇒x+y+2z−z=zx+yz+xyz −zx−yz−2 ∴x+y+z=xyz−2
$$\:\mathrm{1}+{x}=\mathrm{1}+{log}_{{a}} {bc}={log}_{{a}} {a}+{log}_{{a}} {bc} \\ $$$$\mathrm{1}+{x}={log}_{{a}} {abc}\Rightarrow\frac{\mathrm{1}}{\mathrm{1}+{x}}={log}_{{abc}} {a} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{1}+{y}}={log}_{{abc}} {b},\:\:\frac{\mathrm{1}}{\mathrm{1}+{y}}={log}_{{abc}} {c} \\ $$$$\:\:\therefore\:\frac{\mathrm{1}}{\mathrm{1}+{x}}+\frac{\mathrm{1}}{\mathrm{1}+{y}}\:+\frac{\mathrm{1}}{\mathrm{1}+{z}}={log}_{{abc}} {abc}=\mathrm{1} \\ $$$$\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}}\:+\frac{\mathrm{1}}{\mathrm{1}+{y}}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{z}} \\ $$$$\Rightarrow\frac{\mathrm{1}+{y}+\mathrm{1}+{x}}{\mathrm{1}+{y}+{x}+{xy}}=\frac{\mathrm{1}+{z}−\mathrm{1}}{\mathrm{1}+{z}} \\ $$$$\Rightarrow\left(\mathrm{2}+{x}+{y}\right)\left(\mathrm{1}+{z}\right)={z}\left(\mathrm{1}+{x}+{y}+{xy}\right) \\ $$$$\Rightarrow\mathrm{2}+\mathrm{2}{z}+{x}+{xz}+{y}+{yz} \\ $$$$\:\:\:\:\:\:\:\:={z}+{zx}+{yz}+{xyz} \\ $$$$\Rightarrow{x}+{y}+\mathrm{2}{z}−{z}={zx}+{yz}+{xyz} \\ $$$$−{zx}−{yz}−\mathrm{2} \\ $$$$\therefore{x}+{y}+{z}={xyz}−\mathrm{2} \\ $$
Commented by fantastic last updated on 27/Jul/25
thanks sir
$${thanks}\:{sir} \\ $$
Commented by fantastic last updated on 27/Jul/25
best solution I have ever seen Thank you very much
$${best}\:{solution}\:{I}\:{have}\:{ever}\:{seen} \\ $$$${Thank}\:{you}\:{very}\:{much} \\ $$
Commented by som(math1967) last updated on 27/Jul/25
welcome
$${welcome} \\ $$
Answered by fantastic last updated on 27/Jul/25
xyz−2 =(log _a bc)(log _b ca)(log _c ab)−2 =(((log bc)/(log a)))(((log ca)/(log b)))(((log ab)/(log c)))−2 =(((log b+log c)/(log a))×((log c+log a)/(log b))×((log a+log b)/(log c)))−2 =(((log b+log c)/(log b))×((log c+log a)/(log c))×((log a+log b)/(log a)))−2 =(1+((log c)/(log b)))(1+((log a)/(log c)))(1+((log b)/(log a)))−2 ={1+((log a)/(log c))+((log c)/(log b))+(((log c)/(log b))×((log a)/(log c)))}×(1+((log b)/(log a)))−2 =(1+((log a)/(log c))+((log c)/(log b))+((log a)/(log b)))(1+((log b)/(log a)))−2 =(1+((log b)/(log a))+((log a)/(log c))+(((log a)/(log c))×((log b)/(log a)))+((log c)/(log b))+(((log c)/(log b))×((log b)/(log a)))+((log a)/(log b))+(((log a)/(log b))×((log b)/(log a)))−2 =1+((log b)/(log a))+((log a)/(log c))+((log b)/(log c))+((log c)/(log b))+((log c)/(log a))+((log a)/(log b))+1−2 =(((log b)/(log a))+((log c)/(log a)))+(((log c)/(log b))+((log a)/(log b)))+(((log a)/(log c))+((log b)/(log c))) =log _a bc+log _b ca+log _c ab =x+y+z(proved)
$${xyz}−\mathrm{2} \\ $$$$=\left(\mathrm{log}\:_{{a}} {bc}\right)\left(\mathrm{log}\:_{{b}} {ca}\right)\left(\mathrm{log}\:_{{c}} {ab}\right)−\mathrm{2} \\ $$$$=\left(\frac{\mathrm{log}\:{bc}}{\mathrm{log}\:{a}}\right)\left(\frac{\mathrm{log}\:{ca}}{\mathrm{log}\:{b}}\right)\left(\frac{\mathrm{log}\:{ab}}{\mathrm{log}\:{c}}\right)−\mathrm{2} \\ $$$$=\left(\frac{\mathrm{log}\:{b}+\mathrm{log}\:{c}}{\mathrm{log}\:{a}}×\frac{\mathrm{log}\:{c}+\mathrm{log}\:{a}}{\mathrm{log}\:{b}}×\frac{\mathrm{log}\:{a}+\mathrm{log}\:{b}}{\mathrm{log}\:{c}}\right)−\mathrm{2} \\ $$$$=\left(\frac{\mathrm{log}\:{b}+\mathrm{log}\:{c}}{\mathrm{log}\:{b}}×\frac{\mathrm{log}\:{c}+\mathrm{log}\:{a}}{\mathrm{log}\:{c}}×\frac{\mathrm{log}\:{a}+\mathrm{log}\:{b}}{\mathrm{log}\:{a}}\right)−\mathrm{2} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{b}}\right)\left(\mathrm{1}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{c}}\right)\left(\mathrm{1}+\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}\right)−\mathrm{2} \\ $$$$=\left\{\mathrm{1}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{c}}+\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{b}}+\left(\frac{\cancel{\mathrm{log}\:{c}}}{\mathrm{log}\:{b}}×\frac{\mathrm{log}\:{a}}{\cancel{\mathrm{log}\:{c}}}\right)\right\}×\left(\mathrm{1}+\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}\right)−\mathrm{2} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{c}}+\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{b}}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{b}}\right)\left(\mathrm{1}+\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}\right)−\mathrm{2} \\ $$$$=\left(\mathrm{1}+\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{c}}+\left(\frac{\cancel{\mathrm{log}\:{a}}}{\mathrm{log}\:{c}}×\frac{\mathrm{log}\:{b}}{\cancel{\mathrm{log}\:{a}}}\right)+\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{b}}+\left(\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{b}}×\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}\right)+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{b}}+\left(\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{b}}×\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}\right)−\mathrm{2}\right. \\ $$$$=\cancel{\mathrm{1}}+\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{c}}+\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{c}}+\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{b}}+\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{a}}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{b}}\cancel{+\mathrm{1}−\mathrm{2}} \\ $$$$=\left(\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{a}}+\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{a}}\right)+\left(\frac{\mathrm{log}\:{c}}{\mathrm{log}\:{b}}+\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{b}}\right)+\left(\frac{\mathrm{log}\:{a}}{\mathrm{log}\:{c}}+\frac{\mathrm{log}\:{b}}{\mathrm{log}\:{c}}\right) \\ $$$$=\mathrm{log}\:_{{a}} {bc}+\mathrm{log}\:_{{b}} {ca}+\mathrm{log}\:_{{c}} {ab} \\ $$$$={x}+{y}+{z}\left({proved}\right) \\ $$
Commented by fantastic last updated on 27/Jul/25
this is how I used to solve
$${this}\:{is}\:{how}\:{I}\:{used}\:{to}\:{solve} \\ $$

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