Vom folosi urmatoarele formule:
[tex]log_a(xy)=log_ax+log_ay\\
a^{log_ab}=b\\
log_a(x^y)=y\cdot log_ax \rightarrow log_ab \cdot log_bc=log_a(b^{log_bc})=log_ac\\
log_ab\cdot log_ba = 1[/tex]
Conditiile de existenta pentru logaritmi:
a, b, c > 0 si a, b, c ≠ 1
Cum relatia de egalitate este o relatie de echivalenta, putem scrie expresia invers fata a afecta demonstratia:
xyz = x + y + z + 2
Inlocuim:
[tex]xyz=log_a(bc)\cdot log_b(ac)\cdot log_c(ab)=\\
=(log_ab+log_ac)(log_ba+log_bc)(log_ca+log_cb)=\\\\
=(log_ab\cdot log_ba+log_ab\cdot log_bc+log_ac\cdot log_ba+log_ac\cdot log_bc)\cdot\\
\cdot(log_ca+log_cb)=\\\\
=(1+log_a(b^{log_bc})+log_b(a^{log_ac})+log_ac\cdot log_bc)(log_ca+log_cb)\\\\
=(1+log_ac+log_bc+log_ac\cdot log_bc)(log_ca+log_cb)=\\\\
[/tex][tex]=log_ca+log_cb+(log_ac\cdot log_ca)+(log_ac\cdot log_cb)+(log_bc\cdot log_ca)+\\
+(log_bc\cdot log_cb)+(log_ac\cdot log_bc\cdot log_ca) +(log_ac\cdot log_bc \cdot log_cb)=\\\\
=log_c(ab)+1+log_ab+log_ba+1+log_bc+log_ac=\\\\
=log_a(bc)+log_b(ac)+log_c(ab)+2=x+y+z+2[/tex]
