Question

Resolva as equações exponenciais na incógnita x:
a) 3^{2-x} = \frac{1}{27}
b) \sqrt{5} \frac{x}{2} = 125
c) 3 ^{x+1} + 3 ^{x-1} = 810
d) 2 ^{2x} - 2.2 ^{x} - 8 = 0

Answer 1

$a)\ 3^{2-x} = \frac{1}{27} \\ \\3^{2-x} = \Big(\frac{1}{3}\big)^3\\ \\3^{2-x} = 3^{-3}\\ \\2-x=-3\\ \\-3=2-x\\ \\x=2+3\\ \\x=5\\ \\S=\{5\}$
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$\\b)\ \sqrt{5^{\frac{x}{2}}} = 125 \\ \\\big(5^{\frac{x}{2}}\big)^{\frac{1}{2}}=5^3\\ \\5^{\frac{x}{2}\cdot \frac{1}{2} }=5^3\\ \\5^{\frac{x}{4}}=5^3\\ \\\frac{x}{4}=3\\ \\x=4\cdot3\\ \\x=12\\ \\S=\{12\}$

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$c)\ 3^{x+1} + 3^{x-1} = 810 \\ \\3^x\cdot3 + \frac{3^x}{3} = 810 \\ \\3^x\cdot\Big(3 + \frac{1}{3}\Big) = 810 \\ \\3^x\cdot\Big( \frac{9}{3} + \frac{1}{3}\Big) = 810 \\ \\3^x\cdot\Big( \frac{9+1}{3} \Big) = 810 \\ \\3^x\cdot \frac{10}{3} = 810 \\ \\3^x =\frac{3}{10}\cdot 810 \\ \\3^x =243 \\ \\3^x=3^5\\ \\x=5\\ \\S=\{5\}$
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