Solução da equação log2 x + log4 x = 1
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se eu entendi a pergunta, temos:

Então colocando ambos os logs na base 10, ficamos com:
![\frac{ log(x) }{ log(2) } + \frac{ log(x) }{ log(4) } = 1 \\ \\ \frac{ log(x) }{ log(2) } + \frac{ log(x) }{ log( {2}^{2} ) } = 1 \\ \\ \frac{ log(x) }{ log(2) } + \frac{ log(x) }{ 2log(2) } = 1 \\ \\ 2 log(2) log(x) + log(2) log(x) = log(2) \times 2 log(2) \\ \\ log(x) ( \: \: 2 log(2) + log(2) \: \: ) = 2 (log(2) )^{2} \\ \\ log(x) = \frac{2 log(2) }{3} = log( \sqrt[3]{ {2}^{2} } ) \\ \\ log(x) = log( \sqrt[3]{4} ) \\ \\ x = \sqrt[3]{4} \frac{ log(x) }{ log(2) } + \frac{ log(x) }{ log(4) } = 1 \\ \\ \frac{ log(x) }{ log(2) } + \frac{ log(x) }{ log( {2}^{2} ) } = 1 \\ \\ \frac{ log(x) }{ log(2) } + \frac{ log(x) }{ 2log(2) } = 1 \\ \\ 2 log(2) log(x) + log(2) log(x) = log(2) \times 2 log(2) \\ \\ log(x) ( \: \: 2 log(2) + log(2) \: \: ) = 2 (log(2) )^{2} \\ \\ log(x) = \frac{2 log(2) }{3} = log( \sqrt[3]{ {2}^{2} } ) \\ \\ log(x) = log( \sqrt[3]{4} ) \\ \\ x = \sqrt[3]{4}](https://tex.z-dn.net/?f=+%5Cfrac%7B+log%28x%29+%7D%7B+log%282%29+%7D++%2B++%5Cfrac%7B+log%28x%29+%7D%7B+log%284%29+%7D++%3D+1+%5C%5C++%5C%5C++%5Cfrac%7B+log%28x%29+%7D%7B+log%282%29+%7D++%2B++%5Cfrac%7B+log%28x%29+%7D%7B+log%28+%7B2%7D%5E%7B2%7D+%29+%7D++%3D+1+%5C%5C++%5C%5C++%5Cfrac%7B+log%28x%29+%7D%7B+log%282%29+%7D++%2B++%5Cfrac%7B+log%28x%29+%7D%7B+2log%282%29+%7D++%3D+1+%5C%5C++%5C%5C+2+log%282%29++log%28x%29++%2B++log%282%29++log%28x%29++%3D++log%282%29++%5Ctimes+2+log%282%29++%5C%5C++%5C%5C++log%28x%29+%28+%5C%3A++%5C%3A+2+log%282%29++%2B++log%282%29+%5C%3A++%5C%3A++%29+%3D+2+%28log%282%29+%29%5E%7B2%7D+%5C%5C++%5C%5C++log%28x%29++%3D++%5Cfrac%7B2+log%282%29+%7D%7B3%7D+++%3D++log%28+%5Csqrt%5B3%5D%7B+%7B2%7D%5E%7B2%7D+%7D+%29++%5C%5C++%5C%5C++log%28x%29++%3D++log%28+%5Csqrt%5B3%5D%7B4%7D+%29++%5C%5C++%5C%5C+x+%3D++%5Csqrt%5B3%5D%7B4%7D+)
Então colocando ambos os logs na base 10, ficamos com:
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