Question

me ajudem por favor ​

Answer 1

Resposta:

a) x = -1/2   b) x = -1   c) x = 7/2   d) x = 3   e) x = 0   f)  x = 6

Explicação passo-a-passo:

a) $\sqrt{0.1} = \sqrt{10^{-1}}=10^{-1/2}$

   $10^{-\frac{1}{2} }=10^{x}$

   x = -1/2

b) $\frac{27}{125} = \frac{3^3}{5^3} =(\frac{3}{5} )^3 = (\frac{5}{3} )^{-3}$

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

   2x - 1 = -3

   2x = -3 + 1

   2x = -2

   x = -2/2

   x = -1

c) $36=6^2$

   $6^{9-2x}=6^2$

   9 - 2x = 2

   -2x = 2 - 9

   -2x = -7

    x = 7/2

d) $16^{2x -1} = 4^{(2x-1)^{2}}=4^{2(2x-1)}=4^{4x-2}$

   $4^{3x+1}=4^{4x-2}$

   3x + 1 = 4x - 2

   3x - 4x = -2 -1

   -x = -3

    x = 3

e) $\sqrt[7]{21^{x}} = 21^{\frac{x}{7} }$ ; $1 = 21^0$

   $21^0=21^\frac{x}{7}$

   $0 = \frac{x}{7}$

   x = 0

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

  $2^{4x - 2 - (3x+5)}=2^{-1}$

  4x - 2 - (3x + 5) = - 1

  4x -2 -3x - 5 = - 1

  x = - 1 + 2 + 5

  x = 6