What are the solutions to the equation 2^x + 2/2^x = 3?

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Discussion Overview

The discussion centers around solving the equation 2^x + 2/2^x = 3. Participants explore methods to manipulate the equation and derive potential solutions, with a focus on algebraic transformations and the implications of those transformations.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant suggests substituting 2^x with a variable y to simplify the equation.
  • Another participant notes that the equation can be transformed into a quadratic form, leading to a standard quadratic equation.
  • A further contribution details the factorization of the quadratic equation and identifies potential solutions for y, which correspond to values of x.
  • Participants verify the solutions by substituting back into the original equation to check for correctness.

Areas of Agreement / Disagreement

There appears to be agreement on the method of transforming the equation and the resulting solutions, though the initial problem-solving approach and the necessity of showing work are points of discussion.

Contextual Notes

Some participants emphasize the importance of showing prior attempts and where difficulties arise, indicating a collaborative approach to problem-solving.

Who May Find This Useful

This discussion may be useful for individuals seeking to understand algebraic manipulation of exponential equations and the process of solving quadratic equations derived from such transformations.

late6002
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2^x + 2/2^x =3
 
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In future kindly inform what you have tried and where you are stuck so that we can provide steps to proceed

For this put $2^x = y$ and see what you get
 
Hi late6002, welcome to MHB!

For your information, the cited equation can be reduced to quadratic equation...
 
Since this has been here over a month now (and I just can't resist answering):

The equation is 2^x+ 2/2^x= 3. Following Kaliprasad's advice, let y= 2^x. Then the equation becomes, y+ 2/y= 3. Multiply both sides by y to get y^2+ 2= 3y. As anemone said, that is a quadratic equation, y^2- 3y+ 2= 0. And that is easy to factor: (y- 2)(y- 1)= 0. Either y= 2 or y= 1. Since y= 2^x, if y= 2, 2= 2^x so x= 1. If y= 1, 1= 2^x so x= 0.

Check: if x= 0, 2^x= 2^0= 1 so 2^x+ 2/2^x= 1+ 2/1= 1+ 2= 3. If x= 1, 2^x= 2^1= 2 so 2^x+ 2/2^x= 2+ 2/2= 2+ 1= 3.
 

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