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8^(1-x) = 4^(x+2)?

3 Answers

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  • 2 months ago

    Not sure why people are disliking a seemingly correct answer.  I'll do it with an explanation and see if I get what they get:

    8^(1 - x) = 4^(x + 2)

    Both 4 and 8 are powers of 2:

    (2³)^(1 - x) = (2²)^(x + 2)

    The exponent of an exponent is the same as if you took the product of the exponents, so:

    2^[3(1 - x)] = 2^[2(x + 2)]

    We now have the same value on both sides with the same base on both sides so the exponents must also be the same:

    3(1 - x) = 2(x + 2)

    Now we can expand both sides and solve:

    3 - 3x = 2x + 4

    -5x = 1

    x = -1/5

    I do get the same answer as the others.

  • 2 months ago

    8^(1 - x) = 4^(x + 2)

    [2^(3)]^(1 - x) = [2^(2)]^(x + 2) → you konw that: [x^(a)]^(b) = x^(ab)

    2^[3.(1 - x)] = 2^[2.(x + 2)]

    3.(1 - x) = 2.(x + 2)

    3 - 3x = 2x + 4

    - 5x = 1

    x = - 1/5

  • 2 months ago

    (2^3)^(1-x) = (2^2)^(x+2) =>

    2^(3 - 3x) = 2^(2x + 4) =>

    3 - 3x = 2x + 4 =>

    x = -1/5.

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