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Question Number 110183 by bemath last updated on 27/Aug/20

\sqrt{★} \frac{b e}{m a t h} \sqrt{★}

\log_{2} (x) + \log_{3} (x) + \log_{4} (x) = 1

x = ?

Answered by Olaf last updated on 27/Aug/20

\log_{2} x + \log_{3} x + \log_{4} x = 1

\frac{\ln x}{\ln 2} + \frac{\ln x}{\ln 3} + \frac{\ln x}{\ln 4} = 1

\ln x = \frac{1}{\frac{1}{\ln 2} + \frac{1}{\ln 3} + \frac{1}{\ln 4}}

x = e^{1 / (\frac{1}{\ln 2} + \frac{1}{\ln 3} + \frac{1}{\ln 4})} = e^{1 / (\frac{3}{2 \ln 2} + \frac{1}{\ln 3})}

Answered by Aziztisffola last updated on 27/Aug/20

\ln (x) (\frac{1}{\ln 2} + \frac{1}{\ln 3} + \frac{1}{\ln 4}) = 1

\ln (x) (\frac{\ln 3 \ln 4 + \ln 2 \ln 4 + \ln 2 \ln 3}{\ln 2 \ln 3 \ln 4}) = 1

\ln (x) = \frac{\ln 2 \ln 3 \ln 4}{\ln 3 \ln 4 + \ln 2 \ln 4 + \ln 2 \ln 3}

x = e^{\frac{\ln 2 \ln 3 \ln 4}{\ln 3 \ln 4 + \ln 2 \ln 4 + \ln 2 \ln 3}}

Answered by john santu last updated on 28/Aug/20

\log_{2} (x) + \frac{\log_{2} (x)}{\log_{2} (3)} + \frac{\log_{2} (x)}{\log_{2} (4)} = 1

\log_{2} (x) [1 + \frac{1}{\log_{2} (3)} + \frac{1}{2}] = 1

\log_{2} (x) [\frac{3}{2} + \frac{1}{\log_{2} (3)}] = 1

\log_{2} (x) (\frac{\log_{2} (27) + \log_{2} (4)}{\log_{2} (9)}) = 1

\log_{2} (x) (\frac{\log_{2} (108)}{\log_{2} (9)}) = 1

\log_{2} (x) = \frac{\log_{2} (9)}{\log_{2} (108)}

\log_{2} (x) = \log_{108} (9)

\Rightarrow x = 2^{\log_{108} (9)} . △