Question

calcule:

log₂ (1+log₃(1-2x))=2

Answer 1

A equação é:

$\displaystyle log_{2}{[1 + log_{3}{(1 - 2x)}]}= 2$

E para solucioná-la, usaremos as seguintes propriedades:

$\displaystyle 1. \quad \log_{ a }{ c } = b \Rightarrow a^b = c \\ 2. \quad \log_{ a }{ b } + \log_{ a }{ c} = \log_{ a }{ bc } \\ 3. \quad \log_{ a }{ a } = 1$

Assim, temos:

$\displaystyle log_{2}{[1 + log_{3}{(1 - 2x)}]}= 2$

$\displaystyle log_{2}{[log_{3}{3} + log_{3}{(1 - 2x)}]} = 2$

$\displaystyle log_{2}{[log_{3}{3 \cdot (1 - 2x) }]}= 2$

$\displaystyle log_{2}{[log_{3}{(3 - 6x)}]}= 2$

$\displaystyle 2^2 = log_{3}{(3 - 6x)}$

$\displaystyle 4 = log_{3}{(3 - 6x)}$

$\displaystyle 3^4 = 3 - 6x$

$\displaystyle 6x = 3 - 81$

$\displaystyle x = - \frac{78}{6}$

$\displaystyle x =- 13$

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