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suma
: 2 mar 2024, o 09:16
autor: mol_ksiazkowy
Rozwiązać \(\displaystyle{ \log_{2} x + \log_{3} x =1.}\)
Re: suma
: 2 mar 2024, o 09:52
autor: a4karo
Wsk: `x=2^t`
Re: suma
: 2 mar 2024, o 13:35
autor: kerajs
\(\displaystyle{ \log_{2} x + \frac{\log_{2} x}{\log_{2} 3} =\log_{2} 2 \ \ , \ \ x>0 \\
x\cdot x^{\frac{1}{\log_{2} 3}}=2\\
x= \sqrt[1+\frac{1}{\log_{2} 3}]{2}
}\)