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

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The equation 2^x + 2/2^x = 3 can be simplified by substituting y = 2^x, leading to the quadratic equation y^2 - 3y + 2 = 0. This factors to (y - 2)(y - 1) = 0, giving solutions y = 2 and y = 1. Consequently, the corresponding values for x are x = 1 and x = 0. Both solutions satisfy the original equation, confirming their validity. The discussion emphasizes the importance of showing prior attempts when seeking help with mathematical problems.
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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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Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
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