Question

Como resolve essas equações exponenciais?

a) ˣ√2 = 8ˣ

b) (0,01)²ˣ⁻¹ = 1

c) (3/4)ˣ⁺¹ = (64/27)¹⁻²ˣ

d) 3²ˣ -12.3ˣ +27 = 0

e) 4ˣ - 2.2ˣ/5 = 1

f) 100ˣ (0,01)ˣ⁻¹ .(√10)ˣ = 0,001

g) 27ˣ⁺¹ .9ˣ⁻¹ = 3^-1/2

Answer 1

a)
$\sqrt[x]{2} =8^{x}\Rightarrow \left(2\right)^{\dfrac{1}{x}}=8^{x} \Rightarrow \left(2\right)^{\dfrac{1}{x}}=\left(2^{3}\right)^{x} \Rightarrow \left(2\right)^{\dfrac{1}{x}}=2^{3x}\Rightarrow\\ \\ \\ \dfrac{1}{x}=3x \Rightarrow 3x^{2}=1 \Rightarrow x^{2}=\dfrac{1}{3} \Rightarrow x=\pm\sqrt{\dfrac{1}{3}}\Rightarrow x =\pm \dfrac{1}{\sqrt{3}}\\ \\ \\ x=\pm\dfrac{1}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}}\Rightarrow x=\pm \dfrac{\sqrt{3}}{3}$


Ou

$\sqrt[x]{2} =8^{x}\Rightarrow \left(\sqrt[x]{2}\right)^{x}=\left(8^{x}\right)^{x} \Rightarrow 2=8^{x^{2}} \Rightarrow 2=\left(2^{3}\right)^{x^{2}}\Rightarrow\\ \\ \\ 2=2^{3x^{2}}\Rightarrow 3x^{2}=1 \Rightarrow x^{2}=\dfrac{1}{3} \Rightarrow x=\pm\sqrt{\dfrac{1}{3}}\Rightarrow x =\pm \dfrac{1}{\sqrt{3}}\\ \\ \\ x=\pm\dfrac{1}{\sqrt{3}}\cdot \dfrac{\sqrt{3}}{\sqrt{3}}\Rightarrow x=\pm \dfrac{\sqrt{3}}{3}$


b)
$(0,01)^{2x-1}=1\Rightarrow (0,01)^{2x-1}=(0,01)^{0}\Rightarrow 2x-1=0\Rightarrow\\ \\ 2x=1\Rightarrow x= \dfrac{1}{2}$


c)
$\left(\dfrac{3}{4}\right)^{x+1}=\left(\dfrac{64}{27}\right)^{1-2x}\Rightarrow \left(\dfrac{3}{4}\right)^{x+1}=\left(\dfrac{4^{3}}{3^{3}}\right)^{1-2x}\Rightarrow\\ \\ \\ \\ \left(\dfrac{3}{4}\right)^{x+1}=\left[\left(\dfrac{4}{3}}\right)^{3}\right]^{1-2x}\Rightarrow \left(\dfrac{3}{4}\right)^{x+1}=\left[\left(\dfrac{3}{4}}\right)^{-3}\right]^{1-2x}\Rightarrow\\ \\ \\ \\ \left(\dfrac{3}{4}\right)^{x+1}=\left(\dfrac{3}{4}}\right)^{-3\cdot (1-2x)}\Rightarrow$

$x+1 = -3+6x\Rightarrow 5x=4\Rightarrow x= \dfrac{4}{5}$


d)
$3^{2x}-12\cdot 3^{x}+27=0\Rightarrow (3^{x})^{2}-12\cdot 3^{x}+27=0$

Fazemos $3^{x}=y$

$(3^{x})^{2}-12\cdot 3^{x}+27=0\\ \\ y^{2}-12y+27=0\\ \\ \\ y=\dfrac{-b\pm\sqrt{b^{2}-4\cdot a\cdot c}}{2\cdot a}\Rightarrow y=\dfrac{12\pm\sqrt{(-12)^{2}-4\cdot 1\cdot 27}}{2\cdot 1}\\ \\ \\ y=\dfrac{12\pm\sqrt{144-108}}{2}\Rightarrow y=\dfrac{12\pm\sqrt{36}}{2}\Rightarrow y=\dfrac{12\pm 6}{2}\Rightarrow y=\dfrac{\not 12\pm \not 6}{\not 2}\\ \\ \\ y=6\pm3\Rightarrow y_1=6+3=9 \ \ e\ \ y_2=6-3=3$


Agora substituímos os valores de y que encontramos:

$3^{x}=y\Rightarrow 3^{x}=9\Rightarrow 3^{x}=3^{2}\Rightarrow x=2\\ \\ 3^{x}=y\Rightarrow 3^{x}=3\Rightarrow x=1$

Portanto x= 1 ou x = 2.
S = {1,2}


f)
$100^{x}\cdot(0,01)^{x-1}\cdot (\sqrt{10})^{x}=0,001\\ \\ \left(10^{2}\right)^{x} \cdot \left(\dfrac{1}{100}\right)^{x-1}\cdot \left(10^{\frac{1}{2}\right)^{x}=\dfrac{1}{1000}\\ \\ \\ 10^{2x} \cdot \left(\dfrac{1}{10^{2}}\right)^{x-1}\cdot 10^{\frac{x}{2}}=\dfrac{1}{10^{3}}\\ \\ \\ 10^{2x} \cdot \left(10^{-2}\right)^{x-1}\cdot 10^{\frac{x}{2}}=10^{-3}\\ \\ \\ 10^{2x} \cdot 10^{-2x+2}\cdot 10^{\frac{x}{2}}=10^{-3}\\ \\ 10^{2x-2x+2+\frac{x}{2}}=10^{-3}\\ \\ 10^{2+\frac{x}{2}}=10^{-3}$

$2+\dfrac{x}{2}=-3\Rightarrow \dfrac{x}{2}=-3-2\Rightarrow \dfrac{x}{2}=-5\Rightarrow x=-10$


g)
$27^{x+1}\cdot 9^{x-1}=3^{-1/2}\Rightarrow \left(3^{3}\right)^{x+1}\cdot \left(3^{2}\right)^{x-1}=3^{-1/2}\Rightarrow \\ \\ \\ 3^{3x+3}\cdot 3^{2x-2}=3^{-1/2}\Rightarrow 3^{3x+3+2x-2}=3^{-1/2}\Rightarrow\\ \\ \\ 3^{5x+1}=3^{-1/2}\Rightarrow 5x+1=-\dfrac{1}{2}\Rightarrow 5x=-\dfrac{1}{2}-1\Rightarrow\\ \\ \\ 5x=\dfrac{-1-2}{2}\Rightarrow 5x=\dfrac{-3}{2}\Rightarrow x=-\dfrac{3}{10}$