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

find x log(√x)=(√(logx))

$$\:\boldsymbol{\mathrm{find}}\:\boldsymbol{\mathrm{x}} \\ $$$$\:\boldsymbol{\mathrm{log}}\sqrt{\boldsymbol{{x}}}=\sqrt{\boldsymbol{\mathrm{log}{x}}} \\ $$

Answered by mrW2 last updated on 06/Apr/18

x≥0 log x≥0⇒x≥1 (1/2)log x=(√(log x)) ⇒(1/4)(log x)^2 =log x ⇒((1/4)log x−1)log x=0 ⇒log x=0⇒x=1 ⇒(1/4)log x−1=0⇒log x=4⇒x=10^4

$${x}\geqslant\mathrm{0} \\ $$$$\mathrm{log}\:{x}\geqslant\mathrm{0}\Rightarrow{x}\geqslant\mathrm{1} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\mathrm{log}\:{x}=\sqrt{\mathrm{log}\:{x}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{log}\:{x}\right)^{\mathrm{2}} =\mathrm{log}\:{x} \\ $$$$\Rightarrow\left(\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}\:{x}−\mathrm{1}\right)\mathrm{log}\:{x}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{log}\:{x}=\mathrm{0}\Rightarrow{x}=\mathrm{1} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}\:{x}−\mathrm{1}=\mathrm{0}\Rightarrow\mathrm{log}\:{x}=\mathrm{4}\Rightarrow{x}=\mathrm{10}^{\mathrm{4}} \\ $$

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

thanks

$$\mathrm{thanks} \\ $$

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

$$\mathrm{sometimes}\:\mathrm{log}\:\mathrm{means}\:\mathrm{ln},\:\mathrm{everything} \\ $$$$\mathrm{stays}\:\mathrm{the}\:\mathrm{same}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case},\:\mathrm{but} \\ $$$$\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{solution}\:\mathrm{would}\:\mathrm{be}\:{x}={e}^{\mathrm{4}} \\ $$

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

still i don′t get you,more explaination please

$$\boldsymbol{\mathrm{still}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{don}}'\boldsymbol{\mathrm{t}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{you}},\boldsymbol{\mathrm{more}}\:\boldsymbol{\mathrm{explaination}}\:\boldsymbol{\mathrm{please}} \\ $$

Commented by MJS last updated on 06/Apr/18

everything stays the same with logarithm to base b, except the last line: (1/4)log_b x−1=0 ⇒ log_b x=4 ⇒ ⇒ x=b^4 if log=log_(10) ⇒ x=10^4 if log=ln=log_e ⇒ x=e^4

$$\mathrm{everything}\:\mathrm{stays}\:\mathrm{the}\:\mathrm{same}\:\mathrm{with} \\ $$$$\mathrm{logarithm}\:\mathrm{to}\:\mathrm{base}\:{b},\:\mathrm{except}\:\mathrm{the} \\ $$$$\mathrm{last}\:\mathrm{line}: \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\mathrm{log}_{{b}} \:{x}−\mathrm{1}=\mathrm{0}\:\Rightarrow\:\mathrm{log}_{{b}} \:{x}=\mathrm{4}\:\Rightarrow \\ $$$$\Rightarrow\:{x}={b}^{\mathrm{4}} \\ $$$$\mathrm{if}\:\mathrm{log}=\mathrm{log}_{\mathrm{10}} \:\Rightarrow\:{x}=\mathrm{10}^{\mathrm{4}} \\ $$$$\mathrm{if}\:\mathrm{log}=\mathrm{ln}=\mathrm{log}_{{e}} \:\Rightarrow\:{x}={e}^{\mathrm{4}} \\ $$

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

thanx

$$\mathrm{thanx} \\ $$

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