Question 1066900
First, raise e to both sides:
{{{ e^(log((x-2))+log((x+1))) = e^(2*log(2)) }}}

The left hand side becomes  (x-2)(x+1)
The right hand side becomes   {{{ e^(log((2))*2) = 2^2 = 4 }}}

So    {{{ (x-2)(x+1) = 4 }}}
         {{{ x^2 +x -2x -2 = 4 }}}
        {{{  x^2 -x - 6 = 0 }}}
        {{{  (x-3)(x+2) = 0 }}}
    
            x=3  and  x=-2.
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Discard x=-2 because log(-2 - 2) = log(-4) is undefined.
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Check x=3:     log(3-2)+log(3+1) = log(1) + log(4) = 0 + log(4) = log(2^2) = 2log(2)  (ok)
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Answer:   {{{ highlight(x=3) }}}