Question

determine o conjunto verdade log2(2x-1)-log2(x+2)=log2(4x+1)-log2(x+10)

Answer 1

$log\ _2(2x-1)-log\ _2(x+2)=log\ _2(4x+1)-log\ _2(x+10)\\\\\ log\ _2\frac{(2x-1)}{(x+2)} = log\ _2\ \frac{(4x+1)}{(x+10)}\\\\\ (2x-1)(x+10) = (x+2)(4x+1)\\\\\ 2x^2+20x-x-10 = 4x^2+x+8x+2\\\\\ 2x^2-4x^2+19x-9x-12 = 0\\\\ -2x^2+10x-12=0$

$-2x^2+10x-12=0\\\\ x^2-5x+6 = 0\\\\\ \Delta = 25-24\\\\ \Delta = 1\\\\\ x^1 = 3\\\\ x^2 = 2\\\\\ \boxed{S(2,3)}$

Answer 2

Resposta:

log

2

(2x−1)−log

2

(x+2)=log

2

(4x+1)−log

2

(x+10)

log

2

(x+2)

(2x−1)

=log

2

(x+10)

(4x+1)

(2x−1)(x+10)=(x+2)(4x+1)

2x

2

+20x−x−10=4x

2

+x+8x+2

2x

2

−4x

2

+19x−9x−12=0

−2x

2

+10x−12=0

\begin{gathered}-2x^2+10x-12=0\\\\ x^2-5x+6 = 0\\\\\ \Delta = 25-24\\\\ \Delta = 1\\\\\ x^1 = 3\\\\ x^2 = 2\\\\\ \boxed{S(2,3)}\end{gathered}

−2x

2

+10x−12=0

x

2

−5x+6=0

Δ=25−24

Δ=1

x

1

=3

x

2

=2

S(2,3)