Resolva na fórmula por favor as resoluções.... Pergunta de ideia de91293535

emicosonia DEIXAR bases IGUAIS
equação exponencial

1
9×- ¹ = -------- ( 9 = 3x3 = 3²) e (27 = 3x3x3 = 3³)
27

1
(3²)× - ¹ = --------
3³

3²(× - ¹) = 1.3-³
3²× -² = 3-³ ( BASES iguais) (³))

2x - 2 = - 3
2x = - 3 + 2
2x = - 1
x = -1/2 ( resposta)

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robertocarlos5otivr9 1) $9^{x-1}=\dfrac{1}{27}$

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

$3^{2x-2}=3^{-3}$

Igualando os expoentes:

$2x-2=-3~\longrightarrow~2x=-3+2~\longrightarrow~2x=-1~\longrightarrow~\boxed{x=-\dfrac{1}{2}}$

$\text{S}=\{-\frac{1}{2}\}$

2) $(8,32,\dots,2^{37)$

$q=\dfrac{32}{8}=4$

$a_n=a_1\cdot q^{n-1}$

$2^{37}=8\cdot4^{n-1}~\longrightarrow~2^{37}=2^3\cdot(2^2)^{n-1}$

$\dfrac{2^{37}}{2^3}=2^{2n-2}~\longrightarrow~2^{34}=2^{2n-2}$

$2n-2=34~\longrightarrow~2n=36~\longrightarrow~n=\dfrac{36}{2}~\longrightarrow~\boxed{n=18}$

3) $a_1=5$ e $a_6=5120$

$a_6=a_1\cdot q^5$

$5120=5\cdot q^{5}~\longrightarrow~\dfrac{5120}{5}=q^5~\longrightarrow~1024=q^5~\longrightarrow~q=\sqrt[5]{1024}$

$q=\sqrt[5]{(2^2)^5}~\longrightarrow~q=2^2~\longrightarrow~\boxed{q=4}$

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