im supposed to solve this, but im first supposed to convert the bases to be the same and thats what im struggling with​

im supposed to solve this but im first supposed to convert the bases to be the same and thats what im struggling with class=

Respuesta :

9514 1404 393

Answer:

  1

Step-by-step explanation:

From your knowledge of multiplication tables, you know that ...

  8 = 2×2×2 = 2³

  4 = 2×2 = 2²

so the problem can be rewritten as ...

  [tex]\dfrac{8^4}{4^6}=\dfrac{(2^3)^4}{(2^2)^6}=\dfrac{2^{3\cdot4}}{2^{2\cdot6}}=\dfrac{2^{12}}{2^{12}}=\boxed{1}[/tex]

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The applicable rule of exponents is ...

  (a^b)^c = a^(bc)

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Alternate solution

You could use 8^2 = 64 and 4^3 = 64. Then this is (8^2)^2/(4^3)^2 = 64^2/64^2, where the common base is 64, not 2.

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