$\displaystyle \ln x + \ln (x+2) = 4 $ The answer is: $\displaystyle x = -1 + \sqrt{1 + e^4} $ so far I've done: $\displaystyle log_e x + log_e (x+2) = e^4 $ $\displaystyle log_e x^2+2x = e^4 $ Stuck.
Follow Math Help Forum on Facebook and Google+
Originally Posted by desiderius1 $\displaystyle \ln x + \ln (x+2) = 4 $ The answer is: $\displaystyle x = -1 + \sqrt{1 + e^4} $ so far I've done: $\displaystyle log_e x + log_e (x+2) = e^4 $ No it's not $\displaystyle log_e x^2+2x = e^4 $ In the future please put brackets around x^2 + 2x Stuck. $\displaystyle \ln x + \ln (x+2) = 4 $ $\displaystyle \ln | x(x+2)| = 4 $ $\displaystyle x^2 + 2x = e^4 $ $\displaystyle (x+1)^2 = e^4 + 1 $ $\displaystyle x+1 = \sqrt {e^4 + 1} $ $\displaystyle x = -1 + \sqrt{1+e^4} $
Originally Posted by desiderius1 $\displaystyle \ln x + \ln (x+2) = 4 $ The answer is: $\displaystyle x = -1 + \sqrt{1 + e^4} $ so far I've done: $\displaystyle log_e x + log_e (x+2) = e^4 $ That's NOT your orignal equation! $\displaystyle log_e x^2+2x = e^4 $ Stuck.
View Tag Cloud