Question

resuelve las siguientes ecuaciones logarítmica.

a) log(x-9) + log (50x) = 3
b) 2 log (x-1)= 0
c) log 2 x² + 3 log2 x = 10
d) log x + log (x-1) = log (x-2)
e) log6 (2x)-log6 (x+1) = 0

por favor uwu


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Answer 1

Respuesta:

Explicación paso a paso:

a) log(x-9) + log (50x) = 3

   log [(x-9)(50x)] = 3

   (x-9)(50x) = 10³

   (x-9)(50x) = 1000

divide entre 50

(x-9)x = 20

x² - 9x - 20 = 0

sale con la fórmula general

$x = \frac{9+-\sqrt{9^{2}-4(1)(-20)} }{2}$

x = (9±$\sqrt{161}$)/2

pero si la ecuación fuera (9-x)x = 20

9x - x² - 20 = 0

x² - 9x  + 20 = 0

(x - 4)(x - 5)= 0

x = 4    ∨    x= 5

b) 2 log (x-1)= 0

log (x - 1) = log 1

x - 1 = 1

x = 2

c) log 2 x² + 3 log2 x = 10

log2 x² + log2 x³ = 10

$log_{2}$ [x².x³]= $2^{10}$

$x^{5} = 2^{10}$

x = $2^{\frac{10}{5} } =$ 2² = 4

d) log x + log (x-1) = log (x-2)

log [x(x-1)] = log (x-2)

x(x-1) = x-2

x² - x -x + 2 = 0

x² - 2x + 2 = 0

x² - 2x +1 + 1 = 0

(x - 1)² + 1 = 0

(x - 1)² = - 1 no hay solución real

e) log6 (2x)-log6 (x+1) = 0

  $log_{6} \frac{2x}{x+1} = log_{6} 1$

2x/(x+1) = 1

2x = x+1

x = 1