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How do you solve ${log}_{2} (x + 1) - {log}_{2} (x - 1) = 4$ $log_2 (x+1) - log_2 (x-1) = 4$ ?

1 Answer

Dec 27, 2015

$x = \frac{17}{15}$ $x=17/15$

Explanation:

${log}_{2} (x + 1) - {log}_{2} (x - 1) = {log}_{2} (\frac{x + 1}{x - 1})$ $log_2(x+1)-log_2(x-1) = log_2((x+1)/(x-1))$

$2^{4} = 16 \Rightarrow 4 = {log}_{2} (16)$ $2^4 = 16 rArr 4=log_2(16)$

Therefore
$XXX {log}_{2} (x + 1) - {log}_{2} (x - 1) = 4$ $color(white)("XXX")log_2(x+1)-log_2(x-1)=4$
is equivalent to
$XXX {log}_{2} (\frac{x + 1}{x - 1}) = {log}_{2} (16)$ $color(white)("XXX")log_2((x+1)/(x-1))=log_2(16)$

$XXX \frac{x + 1}{x - 1} = 16$ $color(white)("XXX")(x+1)/(x-1) = 16$

$XXX x + 1 = 16 x - 16$ $color(white)("XXX")x+1 = 16x-16$

$XXX - 15 x = - 17$ $color(white)("XXX")-15x = -17$

$XXX x = \frac{17}{15}$ $color(white)("XXX")x=17/15$

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