SOLUTION: I need help on finding all real solutions for 2^2x-24(2^x)=256

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Question 6448: I need help on finding all real solutions for 2^2x-24(2^x)=256
Answer by ichudov(507) About Me  (Show Source):
You can put this solution on YOUR website!
Use y=2%5Ex. 2%5E2x is %282%5Ex%29%5E2, or y%5E2.
y^2-24y-256=0
Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ay%5E2%2Bby%2Bc=0 (in our case 1y%5E2%2B-24y%2B-256+=+0) has the following solutons:

y%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-24%29%5E2-4%2A1%2A-256=1600.

Discriminant d=1600 is greater than zero. That means that there are two solutions: +x%5B12%5D+=+%28--24%2B-sqrt%28+1600+%29%29%2F2%5Ca.

y%5B1%5D+=+%28-%28-24%29%2Bsqrt%28+1600+%29%29%2F2%5C1+=+32
y%5B2%5D+=+%28-%28-24%29-sqrt%28+1600+%29%29%2F2%5C1+=+-8

Quadratic expression 1y%5E2%2B-24y%2B-256 can be factored:
1y%5E2%2B-24y%2B-256+=+1%28y-32%29%2A%28y--8%29
Again, the answer is: 32, -8. Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-24%2Ax%2B-256+%29


so, we have 2^x = 32, -8. x=5 makes 2^x equal to 32. Nothing could make 2^x equal to -8. So the answer is x=5.