solve the equation for x: ln x + ln (x-3) = 0
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Originally Posted by algebra2 solve the equation for x: ln x + ln (x-3) = 0 use the product property of logs ... change to an exponential equation ... solve the quadratic equation ... remember that only values of x > 3 are valid.
ln(x) +l(x-3) =0 ln[x(x-3)] =0 x(x-3) = e^0 x(x-3)=1 x^2 - 3x -1=0 (Use the quadratic formula to find the roots)
Last edited by CaptainBlack; Mar 1st 2009 at 10:35 AM. Reason: removed the last steps, so that the OP has the opertunity to do some of the work
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