Question

Resolva as equações, considerando U=|R:
a) (1/3)^x=9^-²
b) 2^³x-¹=(1/4)^x
c) (√3)^x-²=27
d) 5.2^x=40
e) 3.5^²x-³=15

Answer 1

a) $\left(\dfrac{1}{3}\right)^{x}=9^{-2}$

Note que $\dfrac{1}{3}=3^{-1}$ e $9=3^2$

$(3^{-1})^{x}=(3^2)^{-2}~\longrightarrow~3^{-x}=3^{-4}$

Igualando os expoentes:

$-x=-4~\longrightarrow~\boxed{x=4}$

$\text{S}=\{4\}$


b) $2^{3x-1}=\left(\dfrac{1}{4}\right)^{x}$

Veja que $\dfrac{1}{4}=2^{-2}$

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

$2^{3x-1}=2^{-2x}$

Igualando os expoentes:

$3x-1=-2x~\longrightarrow~3x+2x=1$

$5x=1~\longrightarrow~\boxed{x=\dfrac{1}{5}}$

$\text{S}=\{\frac{1}{5}\}$


c) $(\sqrt{3})^{x-2}=27$

Observe que $\sqrt{3}=3^{\frac{1}{2}$ e $27=3^3$

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

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

Igualando os expoentes:

$\dfrac{x-2}{2}=3~\longrightarrow~x-2=2\cdot3~\longrightarrow~x-2=6$

$x=6+2~\longrightarrow~\boxed{x=8}$

$\text{S}=\{8\}$


d) $5\cdot2^{x}=40$

$2^{x}=\dfrac{40}{5}~\longrightarrow~2^{x}=8$

Note que $8=2^3$

$2^{x}=2^3$

Igualando os expoentes:

$\boxed{x=3}$

$\text{S}=\{3\}$


e) $3\cdot5^{2x-3}=15$

$5^{2x-3}=\dfrac{15}{3}~\longrightarrow~5^{2x-3}=5$

$5^{2x-3}=5^1$

Igualando os expoentes:

$2x-3=1~\longrightarrow~~2x=3+1~\longrightarrow~2x=4$

$x=\dfrac{4}{2}~\longrightarrow~\boxed{x=2}$

$\text{S}=\{2\}$