log2 (x+1) + log2 (x-1) = 3 ..... log to base 2
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Note that $\displaystyle \log_bx+\log_by=\log_b(xy)$ And $\displaystyle \log_bx=y$ if and only if $\displaystyle b^y=x$. With these two propeties of logarithms, you should be able to solve this equation.
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