Question

найти значения выражения log7 21-log7 3+log3 9

Answer 1

Ответ:

Пошаговое объяснение:

ОДЗ x>0 U 9-x>0⇒x<9

x∈(0;9)

log(7)(9-x)-log(21)(9-x)=log(7)(9-x)- lof(7)(9-x)/(1+log(7)3)=

=log(7)(9-x)(1+log(7)3-1)/(1+log(7)3)=log(7)(9-x)*log(7)3/(1+log(7)3)

log(21)x-log(63)x=log(7)x/(1+log(7)3) -log(7)x/(1+2log(7)3)=

=log(7)x(1+2log(7)3-1-log(7)3)/(1+log(7)3)(1+2log(7)3)=

log(7)x*log(7)3/(1+log(7)3)(1+2log(7)3)

[log(7)(9-x)-log(21)(9-x)]/[log(21)x-log(63)x]=

=log(7)(9-x)*log(7)3/(1+log(7)3) :log(7)x*log(7)3/(1+log(7)3)(1+2log(7)3)=

=log(7)(9-x)*log(7)3/(1+log(7)3) * (1+log(7)3)(1+2log(7)3)/l og(7)x*log(7)3=

=log(7)(9-x)*(1+2log(7)3)/log(7)x

log(7)(9-x)*(1+2log(7)3)/log(7)x≤1+2log(7)3

log(7)(9-x)/log(7)x≤1

log(7)(9-x)≤log(7)x

9-x≤x

2x≥9

x≥4,5

Ответ x∈[4,5;9)