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Petrascumpyka
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Cum se determina log5(3x+4)=2

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[tex]log_{5}(3x+4)=2\\ Conditia\ de\ existenta:3x+4>0=>x>- \frac{4}{3}=>x\in( - \frac{4}{3};+\infty)\\ log_{5}(3x+4)=2\cdot log_55\\ log_{5}(3x+4)=\cdot log_55^2\\ log_{5}(3x+4)=\cdot log_525\\ 3x+4=25\\ 3x=21\\ x=7\in( - \frac{4}{3};+\infty)[/tex]

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