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Sсhoolница
@Sсhoolница
July 2022
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Тригонометрическое уравнение. Помогите решить, пожалуйста!
log[3,(3x-6)]^2+log[1/3,(x-2)^5]=5
P.S. На фото понятнее будет
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Answers & Comments
hlopushinairina
Log²₃(3x-6)+log₁/₃(x-2)⁵=5 ОДЗ: x-2>0 x>2
log²₃(3*(x-2))-log₃(x-2)⁵=5
(log²₃3+log₃(x-2)²-5*log₃(x-2)=5
1²+2*log₃(x-2)+log²₃(x-2)-5*log₃(x-2)=5
log²₃(x-2)-3*log₃(x-2)-4=0
log₃(x-2)=t
t²-3t-4=0 D=25
t₁=4 log₃(x-2)=4 x-2=3⁴ x-2=81 x₁=83∈ОДЗ
t₂=-1 log₃(x-2)=-1 x-2=3⁻¹ x-2=1/3 x₂=2¹/₃∈ОДЗ
0 votes
Thanks 0
sangers1959
Verified answer
Log₃(3x-6)²+log₁/₃(x-2)⁵=5;
(x-2)>0;x>2;
2log₃3(x-2)+5log₃⁻¹(x-2)=5;
2log₃3+2log₃(x-2)-5log₃(x-2)=5;
2-3log₃(x-2)=5;
log₃(x-2)=-3;⇒
x-2=1/27;
x=2¹/₂₇;
2 votes
Thanks 1
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Answers & Comments
log²₃(3*(x-2))-log₃(x-2)⁵=5
(log²₃3+log₃(x-2)²-5*log₃(x-2)=5
1²+2*log₃(x-2)+log²₃(x-2)-5*log₃(x-2)=5
log²₃(x-2)-3*log₃(x-2)-4=0
log₃(x-2)=t
t²-3t-4=0 D=25
t₁=4 log₃(x-2)=4 x-2=3⁴ x-2=81 x₁=83∈ОДЗ
t₂=-1 log₃(x-2)=-1 x-2=3⁻¹ x-2=1/3 x₂=2¹/₃∈ОДЗ
Verified answer
Log₃(3x-6)²+log₁/₃(x-2)⁵=5;(x-2)>0;x>2;
2log₃3(x-2)+5log₃⁻¹(x-2)=5;
2log₃3+2log₃(x-2)-5log₃(x-2)=5;
2-3log₃(x-2)=5;
log₃(x-2)=-3;⇒
x-2=1/27;
x=2¹/₂₇;