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If $\log_{2}{(a^{2}-1)}-\log_{2}{(a+1)}=3$, find the value of a.

Answer:

Explanation:

$\log_{2}{(a^2-1)} - \log_{2}{a+1}$

=$\log_{2}{\frac{(a^{2}-1)}{a+1}}$

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

=$\log_{2}{a-1}$

Given, $\log_{2}{a-1}=3$

ie $2^{3} = a-1$

8 = a-1

a = 9