Question

Podem me ajudar na 3??? PorFavor

Answer 1

a)

$3^{x+3} = \frac{1}{81}$

$3^{x+3} = 3^{-4}$

x + 3 = -4

x = -4 - 3

x = -7

b)

$8^{3x+1} = 64$

$8^{3x+1} = 8^2$

3x + 1 = 2

3x = 2 - 1

3x = 1

x = 1/3

c)

$81^{x-2} = \sqrt[4]{27}$

$81^{x-2} = 27^{\frac{1}{4}}$

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

$3^{4{x-8}} = 3^\frac{3}{4}$

4x - 8 = 3/4

4x = 0,75 + 8

4x = 8,75

x = $\frac{8,75}{4}$

x = 2,1875

d)

$\sqrt{16^{x+1}} = \sqrt[4]{64}$

$16^{\frac{x+1}{2}} = 64^{\frac{1}{4}}$

$(2^4)^{\frac{x+1}{2}} = (2^6)^{\frac{1}{4}}$

$2^{\frac{4x+4}{2}} = 2^{\frac{3}{2}}$

$\frac{4x+4}{2} = \frac{3}{2} *(2)$

4x + 4 = 3

4x = 3 - 4

4x = -1

x = $-\frac{1}4}$  ou -0,25

e)

$\sqrt{5^x} * 25^{x+1} = (0,2)^{1-x}$

$5^\frac{x}{2} * (5^2)^{x+1} = (5^{-1})^{1-x}$

$5^\frac{x}{2} * 5^{2x+2} = 5^{-1+x}$

$\frac{x}{2}$ + 2x + 2 = -1 + x

$\frac{x}{2}$ + 2x -x = -1 -2

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

x + 2x = -6

3x = -6

x = $\frac{-6}{3}$

x = -2

f)

$\sqrt[5]{2^x} * \sqrt[3]{4^x} = \sqrt{8^{-x}}$

$\sqrt[5]{2^x} * \sqrt[3]{(2^2)^x} = \sqrt{(2^3)^{-x}}$

$2^{\frac{x}{5}} * \sqrt[3]{2^{2x}} = \sqrt{2^{-3x}}$

$2^{\frac{x}{5}} * 2^{\frac{2x}{3}} = 2^{\frac{-3x}{2}}$

$\frac{x}{5} + \frac{2x}{3} = \frac{-3x}{2}$

mmc(5, 3, 2) = 30

$\frac{6x}{30} + \frac{20x}{30} = \frac{-45x}{30}$

6x + 20x = -45x

26x + 45x = 0

71x = 0

x = 0

g)

$(\frac{2}{5})^{x+3} = (\frac{125}{8})^{x-1} * (0,4)^{2x-3}$

$(\frac{2}{5})^{x+3} = (\frac{5}{2})^{3*{(x-1)}} * (\frac{2}{5} )^{2x-3}$

$(\frac{2}{5})^{x+3} = (\frac{2}{5})^{(-1)*3*{(x-1)}} * (\frac{2}{5} )^{2x-3}$

$(\frac{2}{5})^{x+3} = (\frac{2}{5})^{-3x+3} * (\frac{2}{5} )^{2x-3}$

x+3 = -3x + 3 + 2x - 3

x + 3x - 2x = 3 - 3 - 3

2x = -3

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

h)

$(\frac{1}{27})^{-x} * (3^{3x})^2 = (\frac{1}{3})^{x-1}$

$(\frac{1}{3^3})^{-x} * 3^{6x} = (3^{-1})^{x-1}$

$(\frac{1}{3})^{-3x} * 3^{6x} = 3^{-x+1}$

$(3^{-1})^{-3x} * 3^{6x} = 3^{-x+1}$

$3^{3x} * 3^{6x} = 3^{-x+1}$

3x + 6x = -x + 1

9x + x = 1

10x = 1

x = $\frac{1}{10}$