Question 27481
ln(x)- ln(x+1) = ln(x+3) - ln(x+5)
ln{{{(x/(x+1))}}} = ln{{{((x+3)/(x+5))}}}


and now we get rid of the log. How do we get rid of anything in maths? we do the opposite... so opposite of log is "raise to power". So, doing that, the "raising to the power" and the "log" cancel each other, so we are left with


{{{ x/(x+1) = (x+3)/(x+5) }}}
{{{ x(x+5) = (x+3)(x+1) }}}
{{{ x^2 + 5x = x^2 + 4x + 3 }}}
5x = 4x + 3
x = 3


jon.