find-x-log-x-logx-

Question Number 32912 by mondodotto@gmail.com last updated on 06/Apr/18

find x

\log \sqrt{x} = \sqrt{\log x}

Answered by mrW2 last updated on 06/Apr/18

x ⩾ 0

\log x ⩾ 0 \Rightarrow x ⩾ 1

\frac{1}{2} \log x = \sqrt{\log x}

\Rightarrow \frac{1}{4} {(\log x)}^{2} = \log x

\Rightarrow (\frac{1}{4} \log x - 1) \log x = 0

\Rightarrow \log x = 0 \Rightarrow x = 1

\Rightarrow \frac{1}{4} \log x - 1 = 0 \Rightarrow \log x = 4 \Rightarrow x = 10^{4}

Commented by mondodotto@gmail.com last updated on 06/Apr/18

thanks

Commented by MJS last updated on 06/Apr/18

sometimes \log means \ln, everything

stays the same in this case, but

the 2^{nd} solution would be x = e^{4}

Commented by mondodotto@gmail.com last updated on 06/Apr/18

still i {don}^{'} t get you, more explaination please

Commented by MJS last updated on 06/Apr/18

everything stays the same with

logarithm to base b, except the

last line :

\frac{1}{4} \log_{b} x - 1 = 0 \Rightarrow \log_{b} x = 4 \Rightarrow

\Rightarrow x = b^{4}

if \log = \log_{10} \Rightarrow x = 10^{4}

if \log = \ln = \log_{e} \Rightarrow x = e^{4}

Commented by mondodotto@gmail.com last updated on 07/Apr/18

thanx