Question

Calcule o valor de x em cada item a)log3(x+5)+log3(x−5)+log3 12=1 b)log4(x+2)-log4(x−1)=2

Answer 1

Oie, Td Bom?!

a)

$log_{3}(x + 5) + log_{3}(x - 5) + log_{3}(12) = 1$

$log_{3}((x + 5) \: . \: (x - 5) \: . \: 12) = 1$

$log_{3}((x {}^{2} - 5 {}^{2} ) \: . \: 12) = 1$

$log_{3}((x {}^{2} - 25) \: . \: 12 ) = 1$

$log_{3}(12x {}^{2} - 300 ) = 1$

$12x {}^{2} - 300 = 3 {}^{1}$

$12x {}^{2} - 300 = 3$

$4x {}^{2} - 100 = 1$

$4x {}^{2} = 1 + 100$

$4x {}^{2} = 101$

$x {}^{2} = \frac{101}{4}$

$x = \sqrt{ \frac{101}{4} }$

$x = \frac{ \sqrt{101} }{2}$

b)

$log_{4}(x + 2) - log_{4}(x - 1) = 2$

$log_{4}( \frac{x + 2}{x - 1} ) = 2$

$\frac{x + 2}{x - 1} = 4 {}^{2}$

$\frac{x + 2}{x - 1} = 16$

$(x - 1) \: . \: \frac{x + 2}{x - 1} = 16(x - 1)$

$x + 2 = 16(x - 1)$

$x + 2 = 16x - 16$

$x - 16x = - 16 - 2$

$- 15x = - 18$

$x = - 18 \div ( - 15)$

$x = \frac{18}{15} \div (3)$

$x = \frac{6}{5}$

Att. Makaveli1996