log-2-x-log-3-x-1-x-

Question Number 131890 by Study last updated on 09/Feb/21

{l o g}_{2} x + {l o g}_{3} x = 1 x = ?

Answered by Raxreedoroid last updated on 09/Feb/21

\frac{{l o g}_{3} x}{{l o g}_{3} 2} + {l o g}_{3} x = 1

{l o g}_{3} \sqrt{x^{2}} = 1

x = 3

Commented by liberty last updated on 09/Feb/21

wrong

Commented by Raxreedoroid last updated on 09/Feb/21

ah sorry a mistake

\frac{{l o g}_{3} x}{{l o g}_{3} 2} + {l o g}_{3} x = 1

{l o g}_{3} x \cdot (1 + {l o g}_{3} 2) = 1

{l o g}_{3} x = \frac{1}{1 + {l o g}_{3} 2}

x = 3^{\frac{1}{1 + {l o g}_{3} 2}} \approx 1.9613 \dots

Commented by EDWIN88 last updated on 09/Feb/21

it should be \Rightarrow \log_{3} (x) [1 + \frac{1}{\log_{3} (2)}] = 1

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

\log_{3} (x) = \frac{\log_{3} (2)}{\log_{3} (6)} \Rightarrow x = 3^{\log_{6} (2)} \approx 1.529592

3^{\log_{6} 2}

1.529592

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