find x log(log2+log


Question and Answers Forum

All Questions Topic List
None Questions
Previous in All Question Next in All Question
Previous in None Next in None
Question Number 35743 by mondodotto@gmail.com last updated on 22/May/18

$find x$ $\log (\log 2 + \log_{2} (x + 1)) = 0$
Commented by prof Abdo imad last updated on 25/May/18
$if log mean ln (e)⇔ ln(2) +ln_2 (x+1) =1 ⇔ ln(2) + ((ln(x+1))/(ln(2))) =1 with x>−1 ⇔{ln(2)}^2 +ln(x+1) = ln(2) ⇔ln(x+1)=ln(2) −{ln(2)}^2 ⇔ x+1 =e^(ln(2) −(ln(2))^2 ) =2 e^(−{ln(2)}^2 ) ⇔ x = 2 e^(−{ln(2)}^2 ) −1 .$
$i f l o g m e a n l n (e) \Leftrightarrow l n (2) + {l n}_{2} (x + 1) = 1$ $\Leftrightarrow l n (2) + \frac{l n (x + 1)}{l n (2)} = 1 w i t h x > - 1$ $\Leftrightarrow {l n (2)}^{2} + l n (x + 1) = l n (2)$ $\Leftrightarrow l n (x + 1) = l n (2) - {l n (2)}^{2} \Leftrightarrow$ $x + 1 = e^{l n (2) - {(l n (2))}^{2}} = 2 e^{- {l n (2)}^{2}} \Leftrightarrow$ $x = 2 e^{- {l n (2)}^{2}} - 1 .$
	Answered by MJS last updated on 23/May/18

	$\log = \ln I hope . . .$ $\ln (\ln (2) + \log_{2} (x + 1)) = 0$ $\ln (2) + \log_{2} (x + 1) = 1$ $\log_{2} (x + 1) = 1 - \ln (2)$ $(x + 1) = 2^{1 - \ln (2)}$ $x = \frac{2}{2^{\ln (2)}} - 1 \approx . 237$
	Commented by mondodotto@gmail.com last updated on 23/May/18

	$thanx$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum