f(x)=2^log[(2x-3)(x+2)] f(x)=2 tolong caranya
Matematika
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Pertanyaan
f(x)=2^log[(2x-3)(x+2)] f(x)=2 tolong caranya
2 Jawaban
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1. Jawaban Radesu
F(x)=2^log[(2x-3)(x+2)] f(x)=2
F(x)=2^ -
2. Jawaban Takamori37
[tex]^2\log{(2x-3)(x+2)}=2 \\ ^2\log{(2x-3)(x+2)}=^2\log{4} \\ (2x-3)(x+2) = 4 \\ 2x^2+4x-3x-6=4 \\ 2x^2+x-6-4=0 \\ 2x^2+x-10=0 \\ (2x+5)(x-2)=0 \\ \displaystyle x=-\frac{5}{2}$ dan $x=2[/tex]