Question 1067887
{{{ log(( 1-x )) - log(( 2+x )) = 2 }}}
I assume the base of the logs is {{{ 10 }}}
Use the substitution {{{ 2 = log( 100 ) }}}
{{{ log( ((1-x)/(2+x)) ) = log(100) }}}
{{{ ( 1-x )/( 2+x ) = 100 }}}
{{{ 1 - x = 100*( 2+x ) }}}
{{{ 1 - x = 200 + 100x }}}
{{{ 101x = -199 }}}
{{{ x = -1.970297 }}}
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check answer:
{{{ log(( 1-x )) - log(( 2+x )) = 2 }}}
{{{ log(( 1-(-1.970297 )) - log(( 2+1.970297 ))) = 2 }}}
{{{ log(( 2.970297 )) - log(( .02970297 )) = log(( 100 )) }}}
{{{ log(( 2.970297/.02970297 )) = log(( 100 )) }}}
{{{ log(( 99.99999899 )) = log(( 100 )) }}}
OK
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