Question and Answers Forum

All Questions      Topic List

Logarithms Questions

Previous in All Question      Next in All Question      

Previous in Logarithms      Next in Logarithms      

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

$${this}\:{is}\:{how}\:{I}\:{used}\:{to}\:{solve} \\ $$

Terms of Service

Privacy Policy

Contact: [email protected]