Question

$log4(x + 3) - log4(x - 1) = 2 - log_{4}(8)$

Answer 1

Hi Friend !!!

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$log_{4}(x + 3) - log_{4}(x - 1) = 2 - log_{4}(8) \\ \\ log_{4}( \frac{x + 3}{x - 1} ) = log_{4}(16) - log_{4}(8) \\ \\ log_{4}( \frac{x + 3}{x - 1} ) = log_{4}( \frac{16}{8} ) \\ \\ \frac{x + 3}{x - 1} = \frac{16}{8}$

8x+24 = 16x-16

24+16 = 16x-8x

40 = 8x

x = 5

Hope it helps u:-)

Answer 2

HELLO.......FRIEND!!

THE ANSWER IS HERE,

$= > \: log_{4}(x + 3) - log_{4}(x - 1) = 2 - log_{4}(8)$

$= > \: log_{4}( \frac{x + 3}{x - 1} ) = log_{4}(16) - log_{4}(8)$

$= > \: log_{4}( \frac{x +3 }{x - 1} ) = log_{4}( \frac{16}{8} )$

$= > \: \frac{x + 3}{x - 1} = \frac{16}{8} = 2$

$= > \: x + 3 = 2(x - 1)$

$= > \: x + 3 = 2x - 2$

$= > \: \: x = 5$


:-)Hope it helps u.