Question

Alguém pode me ajudar a responder as questões da imagem ?

Answer 1

G)

$\sqrt{ {5}^{x - 1}}. \sqrt[x]{ {5}^{4x - 10} } - \sqrt[2x]{ {5}^{3x - 1} } = 0$

${5}^{ \frac{x - 1}{2} } . {5}^{ \frac{4x - 10}{x}} - {5}^{ \frac{3x - 1}{2x} } = 0 \\ {5}^{ \frac{x - 1}{2} } . {5}^{ \frac{4x - 10}{x} } = {5}^{ \frac{3x - 1}{2x} }$

${5}^{ \frac{x(x - 1) + 2(4x - 10)}{2x}} = {5}^{ \frac{3x - 1}{2x} }$

$\frac{x(x - 1) + 2(4x - 10)}{2x} = \frac{3x - 1}{2x}$

$x(x - 1) + 2(4x - 10) = 3x - 1 \\ {x}^{2} - x + 8x - 20 - 3x + 1 = 0 \\ {x}^{2} + 4x - 19 = 0$

∆=16+76=92

x=(-4±2√23)/2

x'=(-4+2√23)/2=2(-2+√23)/2

x'=-2+√23

H)

${ ({2}^{x}) }^{x + 4} = 32 \\ {2}^{ {x}^{2} + 4x} = {2}^{5} \\ {x}^{2} + 4x = 5 \\ {x}^{2} + 4x - 5 = 0$

${x}^{2} - x + 5x - 5 = 0 \\ x(x - 1) + 5(x - 1) = 0 \\ (x - 1)(x + 5) = 0$

$x - 1 = 0 \\ x = 1 \\ x + 5 = 0 \\ x = - 5$

i)

${2}^{3x - 1}. {4}^{2x + 3} = {8}^{3x} \\ {2}^{3x - 1}. {2}^{4x + 6} = {2}^{9} \\ {2}^{7x + 5} = {2}^{9} \\ 7x + 5 = 9$

$7x =9 - 5 \\ 7x = 4 \\x = \frac{4}{7}$

2)

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

${9}^{y} = {3}^{x - 9} \\ {3}^{2y} = {3}^{x - 9} \\ 2y = x - 9 \\ x = 2y + 9$

$3y + 3 = 2y + 9 \\ 3y - 2y = 9 - 3 \\ y = 6$

$x = 2y + 9 \\ x = 2.6 + 9 \\ x = 12 + 9 \\ x = 21$

$x + y = 21 + 6 = 27$

3)

${2}^{2x - 3} - 3. {2}^{x - 1} + 4 = 0 \\ \frac{ { ({2}^{x}) }^{2} }{ {2}^{3} } - 3. \frac{ {2}^{x} }{2} + 4 = 0$

$Faça \: {2}^{x} = k$

$\frac{ {k}^{2} }{8} - 3. \frac{k}{2} + 4 = 0 \times (8) \\ {k}^{2} - 12k + 32 = 0$

${k}^{2} - 4k - 8k + 32 = 0 \\ k(k - 4) - 8(k - 4) = 0 \\ (k - 4)(k - 8) = 0$

$k - 4 = 0 \\ k = 4 \\ k - 8 = 0 \\ k = 8$

${2}^{x} = k \\ {2}^{x}= 4 \\ {2}^{x} = {2}^{2} \\ x = 2$

${2}^{x} = k \\ {2}^{x} = 8 \\ {2}^{x} = {2}^{3} \\ x = 3$

4)

a)

${3}^{x + 2} + {3}^{x - 1} = 84 \\ {3}^{x} . {3}^{2} + {3}^{x}. {3}^{ - 1} = 84 \\ {3}^{x} (9 + \frac{1}{3} ) = 84 \\ {3}^{x} . \frac{28}{3} = 84$

${3}^{x} = \frac{3}{28} .84 \\ {3}^{x} = 9 \\ {3}^{x} = {3}^{2} \\ x = 2$

b)

${64}^{x} = 256 \\ {2}^{6x} = {2}^{8} \\ 6x = 8 \\x = \frac{8}{6} = \frac{4}{3}$

c)

${2}^{x + 1} + {2}^{x - 1} = 20 \\ {2}^{x} .2 + {2}^{x} . {2}^{ - 1} = 20 \\ {2}^{x}(2 + {2}^{ - 1} ) = 20 \\ {2}^{x}(2 + \frac{1}{2}) = 20$

${2}^{x} . \frac{5}{2} = 20 \\ {2}^{x} = 20. \frac{2}{5} \\ {2}^{x} = 8 \\ {2}^{x} = {2}^{3} \\ x = 3$

d)

${3}^{x + 1} - {3}^{x + 2} = - 54 \\ {3}^{x} .3 - {3}^{x} . {3}^{2} = - 54 \\ {3}^{x}(3 - 9) = - 54 \\ {3}^{x} .( - 6) = - 54$

${3}^{x} = \frac{ - 54}{ - 6} \\ {3}^{x} = 9 \\ {3}^{x} = {3}^{2} \\ x = 2$

e)

${2}^{x + 1} + {2}^{1 - x} - 5 = 0$

${2}^{x}.2 + 2. { ({2}^{x}) }^{ - 1} - 5 = 0 \\ Faça \: {2}^{x} = y$

$2y + \frac{2}{y} - 5 = 0 \times y \\ 2 {y}^{2} + 2 - 5y = 0 \\ 2 {y}^{2} - 5y + 2 = 0$

$2 {y}^{2} -4 y - y + 2 = 0 \\ 2y(y - 2) - 1(y - 2) = 0 \\ (y - 2)(2y - 1) = 0$

$y - 2 = 0 \\ y = 2 \\ 2y - 1 = 0 \\ 2y = 1 \\ y = \frac{1}{2}$

${2}^{x} = y \\ {2}^{x} = 2 \\ x = 1$

${2}^{x} = y \\ {2}^{x} = \frac{1}{2} \\ {2}^{x} = {2}^{ - 1} \\ x = - 1$

S={-1,1}