Question

Resolva a equação$(0,25)^{x-1}$ = $(\frac{1}{8}) ^{1-x}$

Answer 1

Antes de tudo, vamos estabelecer algumas propriedades:

$1. \quad \displaystyle a^{-b} = \left( \frac{ 1 }{ a }^b \right) \\ 2. \quad (a^b )^c = a^{bc} \\ 3. \quad Se \, \, a^b = a^c, logo \, \, b = c$

Prosseguimos:

$\displaystyle (0,25)^{x-1} = \left( \frac{1}{8} \right) ^{1-x}$

$\displaystyle \left(\frac{1}{4}\right)^{x-1} = \left( \frac{1}{8} \right) ^{1-x}$

$\displaystyle 4^{-(x-1)} = 8^{-(1-x)}$

$\displaystyle 4^{-x+1} = 8^{-1+x}$

$\displaystyle (2^2)^{-x+1} = (2^3) ^{-1+x}$

$\displaystyle 2^{2 \cdot (-x+1)} = 2^{3 \cdot (-1+x)}$

$\displaystyle 2^{-2x+2} = 2^{-3+3x}$

$\displaystyle -2x+2 = -3+3x$

$\displaystyle 2 + 3 = 3x + 2x$

$\displaystyle 5 = 5x$

$\displaystyle 5x = 5$

$\displaystyle x = \frac{ 5}{ 5 }$

$\displaystyle x = 1$

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