persamaan logaritma 1. 7 log (x²-4x-4)= 3 log (x²-4x-4) 2. log (x²-2x+7) = log (3x+1)
Matematika
reztu
Pertanyaan
persamaan logaritma
1. 7 log (x²-4x-4)= 3 log (x²-4x-4)
2. log (x²-2x+7) = log (3x+1)
1. 7 log (x²-4x-4)= 3 log (x²-4x-4)
2. log (x²-2x+7) = log (3x+1)
1 Jawaban
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1. Jawaban anggiresti
1. nomor 1 itu soalnya kayak gini 7 log (x²-4x-4) = 3 log (x²-4x-4) atau ^7 log .... = ^3 log ....???
2. log (x²-2x+7) = log (3x+1)
f(x) = g(x)
x²-2x+7 = 3x+1
x²-5x+6 = 0
(x-2)(x-3) = 0
x-2=0 x-3=0
x=2 x=3