How do i solve this algebraically?

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

How do i know this is true? Are there any tricks?

Update:

What Im trying to say is that how do i know that

-(2^-2) + 2^(-3)= 2^(-3). without solving it?

Are there any tricks I can do to manipulate the exponents

5 Answers

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  • 6 years ago

    The expression is always true for any x. Expand the leftmost term and then combine terms and you will see.

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

    .. -(2^1)(2^(-x)) + 2^(-x) = -2^(-x) … left term expanded

    .. -2*2^(-x) + 2^(-x) = -2^(-x) … evaluate 2^1

    .. 2^(-x)*(-2 + 1) = -2^(-x) … factor out 2^(-x) on the left

    .. -2^(-x) = -2^(-x) … simplify

    _____

    Your example dropped a negative sign on the right. Put that back in and you effectively have

    .. -1/4 + 1/8 = -1/8 … true

  • 6 years ago

    a rule for exponents ___ a^b * a^c = a^(b + c)

    - 2^(1 - x) = - 2^1 * 2^(-x)] = -2 * 2^(-x)

    adding 2 * 2^(-x) to the original equation (on both sides) ___ 2^(-x) = 2^(-x)

    so the equation is true ; unfortunately, x can have any value

  • -(2^(1 - x)) + 2^(-x) = -2^(-x)

    -(2^1 / 2^x) + 1 / 2^x = -1 / 2^x

    (-2 + 1) / 2^x = -1 / 2^x

    -1 / 2^x = -1 / 2^x

  • 6 years ago

    - 2^(1-x) + 2^(-x) =-2^(-x) now multiply through by 2^x, so

    -2^1 + 2^0=-2^0, i.e.

    -2+1=-1 which is independent of x, so x can have any value.

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  • 6 years ago

    - 2^(1-x) + 2^(-x) = -2^(-x),

    - 2^(1-x) + 2^(-x) + 2^(-x) = 0,

    - 2^(1-x) + 2*2^(-x) = 0,

    OR,

    - 2^(1-x) + 2^(1-x) = 0,

    OR,

    0 = 0,

    Hence,

    NO Solution ....... >===============< ANSWER

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