Respuesta :

Nayefx

Answer:

[tex] \large1) \sf {2}^{8x} = {2}^{5 x+ 15} [/tex]

[tex] \large \: 2)\sf {2}^{ - 4x} = {2}^{2x + 2} [/tex]

Step-by-step explanation:

to understand this

you need to know about:

  • law of exponent
  • PEMDAS

let's solve:

  1. [tex] \sf rewrite \: 16 \: as \: {2}^{4} : \\ \sf {2}^{4(2x)} = {32}^{x + 3} [/tex]
  2. [tex] \sf rewrite \: 32 \: as \: {2}^{5} : \\ {2}^{4(2x)} = {2}^{5(x + 3)} [/tex]
  3. [tex] \sf distribute : \\ \sf {2}^{8x} = {2}^{5 x+ 15} [/tex]

let's figure out the second one:

  1. [tex] \sf rewrite \: \frac{1}{16} \: as \: {2}^{ - 4} : \\ \sf {2}^{ - 4(x)} = {4}^{x + 1} [/tex]
  2. [tex] \sf rewrite \: 4 \: as \: {2}^{2} : \\ \sf {2}^{ - 4(x)} = {2}^{2(x + 1)} [/tex]
  3. [tex] \sf distribute : \\ {2}^{ - 4x} = {2}^{2x + 2} [/tex]

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