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.
 

Thread 'Area of Overlapping Squares'

Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?
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