http://img19.imageshack.us/img19/711...c4bdfa6aa2.png

what would the x be? i am so confused...

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- Oct 2nd 2009, 02:39 PMbadshahkhanI can't figure out this log problem!!!!
http://img19.imageshack.us/img19/711...c4bdfa6aa2.png

what would the x be? i am so confused... - Oct 2nd 2009, 02:56 PMpickslides
$\displaystyle \log_2x+ \log_2(x+3) = \log_28$

addition of logs gives

$\displaystyle \log_2(x(x+3)) = \log_28$

$\displaystyle \log_2(x^2+3x) = \log_28$

raising both sides to the power of 2

$\displaystyle x^2+3x = 8$

$\displaystyle x^2+3x - 8=0$

now we have a quadratic, do you know how to solve this? - Oct 2nd 2009, 03:09 PMbadshahkhan
yes ofcourse! 1.7 i believe

- Oct 2nd 2009, 04:18 PMmr fantastic