Question

Qual o conjunto solução de: a) Log₁₀√ˣ + log₁₀₀ˣ = 2 b) log₂ˣ - 9.log₈ˣ = 4

Answer 1

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$a) \\ \log_{10} \sqrt{x} +\log_{100}x=2 \\ \\ mudar~~\log_{100}x~~para~base~~10 \\ \\ \log_{10} \sqrt{x} + \frac{\log_{10}x}{\log_{10}100} =2 \\ \\ \log_{10} \sqrt{x} + \frac{\log_{10}x}{2} =2 \\ \\ mmc=2 \\ \\ 2\log_{10} \sqrt{x} +\log_{10}x=4 \\ \\ \log_{10}( \sqrt{x} )^2+\log_{10}x=4 \\ \\ \log_{10}x+\log_{10}x=4 \\ \\ 2\log_{10}x=4 \\ \\ \log_{10}x=4\div2 \\ \\ \log_{10}x=2 \\ \\ x=10^2 \\ \\ x=100$
   

$b) \\ \log_2x-9\log_8x=4 \\ \\ mudar~~\log_8x~~para~~base~~2 \\ \\ \log_2x- 9\frac{\log_2x}{\log_28} =4 \\ \\ \log_2x-\not9^3 \frac{\log_2x}{\not3} =4 \\ \\ \log_2x-3\log_2x=4 \\ \\ -2\log_2x=4 \\ \\ \log_2x=4\div(-2) \\ \\ \log_2x=-2 \\ \\ x=2^{-2} \\ \\ x= \frac{1}{2^2} = \frac{1}{4}$