(log по 5 (25х^2) + 48) / (log по 5 от х^2 - 49) - вопрос №863424952524852491 от dolbab 30.07.2022 05:50

Question

Алгебра: (log по 5 (25х^2) + 48) / (log по 5 от х^2 - 49) = -1

dolbab · Answer

Log(5)(25x²)+48=log(5)25+log(5)x²+48=2+2log(5)x+48=log(5)x+50
log(5)x²-49=2log(5)x-49
----------------------------------------------
x>0
(2log(5)x+50)/(2log(5)x-49)≥-1
(2log(5)x+50)/(2log(5)x-49)+1≥0
(2log(5)x+50+2log(5)x-49)/(2log(5)x-49)≥0
(4log(5)x+1)/(2log(5)x-49)≥0
log(5)x=t
(4t+1)/(2t-49)≥0
t=-1/4 t=49/2
+ _ +
-----------[-1/4]----------------(49/2)-------------
t≤-1/4⇒log(5)x≤-1/4⇒ $x \leq 1/ \sqrt[4]{5}$
t>49/2⇒log(5)x>49/2⇒x> $5 ^{24} \sqrt{5}$
ответ ∈(0; $1/ \sqrt[4]{5}$ ] U ( $5 ^{24} \sqrt{5}$ ;∞)

(log по 5 (25х^2) + 48) / (log по 5 от х^2 - 49) > = -1