Question

resolva as seguintes equações exponenciais

Answer 1

a) $9^{x+4}=27$

Note que $9=3^2$ e $27=3^3$. Substituindo:

$(3^2)^{x+4}=3^3$

$3^{2x+8}=3^3$

Igualando os expoentes:

$2x+8=3$

$2x=3-8~\longrightarrow~2x=-5~\longrightarrow~\boxed{x=-\dfrac{5}{2}}$

$\text{S}=\left\{-\dfrac{5}{2}\right\}$


b) $32^{x}=\dfrac{1}{\sqrt[5]{4}}$

Observe que $32=2^5$ e $\dfrac{1}{\sqrt[5]{4}}=\dfrac{1}{\sqrt[5]{2^2}}=\dfrac{1}{2^{\frac{2}{5}}}=2^{-\frac{2}{5}}}$

Substituindo:

$(2^5)^{x}=2^{-\frac{2}{5}}$

$2^{5x}=2^{-\frac{2}{5}}$

Igualando os expoentes:

$5x=-\dfrac{2}{5}$

$25x=-2~\longrightarrow~\boxed{x=-\dfrac{2}{25}}$

$\text{S}=\left\{-\dfrac{2}{25}\right\}$


c) $5^{2x+1}=\dfrac{1}{625}$

Veja que $\dfrac{1}{625}=\dfrac{1}{5^4}=5^{-4}$. Substituindo:

$5^{2x+1}=5^{-4}$

Igualando os expoentes:

$2x+1=-4~\longrightarrow~2x=-5~\longrightarrow~\boxed{x=-\dfrac{5}{2}}$

$\text{S}=\left\{-\dfrac{5}{2}\right\}$


d) $5^{x^2+2x}=1$

Note que $5^{0}=1$. Substituindo:

$5^{x^2+2x}=5^{0}$

Igualando os expoentes:

$x^2+2x=0$

$x\cdot(x+2)=0$

$\boxed{x'=0}$

$x+2=0~\longrightarrow~\boxed{x"=-2}$

$\text{S}=\{-2,0\}$