# I can't figure out this log problem!!!!

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• October 2nd 2009, 03:39 PM
badshahkhan
I can't figure out this log problem!!!!
http://img19.imageshack.us/img19/711...c4bdfa6aa2.png

what would the x be? i am so confused...
• October 2nd 2009, 03:56 PM
pickslides
$\log_2x+ \log_2(x+3) = \log_28$

addition of logs gives

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

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

raising both sides to the power of 2

$x^2+3x = 8$

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

now we have a quadratic, do you know how to solve this?
• October 2nd 2009, 04:09 PM
badshahkhan
yes ofcourse! 1.7 i believe
• October 2nd 2009, 05:18 PM
mr fantastic
Quote:

Originally Posted by badshahkhan
yes ofcourse! 1.7 i believe

If an answer correct to 1 decimal place is required then this is correct. But if an exact answer is required then it's not correct.