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# How do i solve this algebraically?

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

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

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

- SqdancefanLv 76 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

- scott8148Lv 76 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

- 6 years ago
-(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

- Wonder WhoLv 66 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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- Fazaldin ALv 76 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