Question

Resolva as equações exponenciais:
a) $2x {}^{ + 1} + 2 {}^{x + 2} + 2 {}^{x + 3} = 224$
b) $3 {}^{3x - 1} \times 9 {}^{2x + 3} = 27 {}^{3 - x}$
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Answer 1

a)

$2^{x+1} + 2^{x+2} + 2^{x+3} = 224\\2^{x+1} + 2^{x+1} \cdot 2 + 2^{x+1} \cdot 2^2 = 224\\2^{x+1}(1 + 2 + 2^2) = 7 \cdot 32\\2^{x+1} \cdot 7= 7 \cdot 2^5\\2^{x+1} = 2^5\\x + 1 = 5\\x = 4$

b)

$3^{3x-1} \cdot 9^{2x+3} = 27^{3-x}\\3^{3x-1} \cdot (3^2)^{2x+3} = (3^3)^{3-x}\\3^{3x-1} \cdot 3^{2(2x+3)} = 3^{3(3-x)}\\3^{3x-1} \cdot 3^{4x+6} = 3^{9 - 3x}\\3^{3x-1 + 4x + 6} = 3^{9 - 3x}\\3^{7x + 5} = 3^{-3x + 9}\\7x + 5 = -3x + 9\\10x = 4\\x = 0,4$