simelaalbo
simelaalbo simelaalbo
  • 28-09-2020
  • Mathematics
contestada

Solve for x given that
[tex] {64}^{2x - 1} = {32}^{x + 1} [/tex]
​

Respuesta :

jimrgrant1 jimrgrant1
  • 28-09-2020

Answer:

x = [tex]\frac{11}{7}[/tex]

Step-by-step explanation:

note that 64 = [tex]2^{6}[/tex] and 32 = [tex]2^{5}[/tex] , thus

[tex](2^{6}) ^{2x-1}[/tex] = [tex](2^{5}) ^{x+1}[/tex] , that is

[tex]2^{6(2x-1)}[/tex] = [tex]2^{5(x+1)}[/tex]

[tex]2^{12x-6}[/tex] = [tex]2^{5x+5}[/tex]

Since the bases on both sides are equal, then equate the exponents

12x - 6 = 5x + 5 ( subtract 5x from both sides )

7x - 6 = 5 ( add 6 to both sides )

7x = 11 ( divide both sides by 7 )

x = [tex]\frac{11}{7}[/tex]

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