find-x-log-log2-log-2-x-1-0-

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

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 \dots

\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