Solve:

$l o g_{5} (x + 1) - 1 = 1 + l o g_{5} (x - 1)$

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Solution

$l o g_{5} (x + 1) - 1 = 1 + l o g_{5} (x - 1)$
$l o g_{5} (x + 1) - l o g_{5} (x - 1) = 2$
$l o g_{5} (\frac{x + 1}{x - 1}$ ) = $2$

$l o g_{5} (\frac{x + 1}{x - 1}$ ) = $2 l o g_{5} 5$

$l o g 5 (\frac{x + 1}{x - 1}$ ) = $l o g_{5} 25$

( $\frac{x + 1}{x - 1}$ ) = 25

(x+1) = 25 (x-1)

x+1 = 25x - 25

26 = 24 x

x = $\frac{26}{24}$

x = $\frac{13}{12}$

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