document.write( "Question 6448: I need help on finding all real solutions for 2^2x-24(2^x)=256 \n" ); document.write( "

Algebra.Com's Answer #3496 by ichudov(507) $\"\"$ $\"About$

You can put this solution on YOUR website!
Use $\"y=2%5Ex\"$ . $\"2%5E2x\"$ is $\"%282%5Ex%29%5E2\"$ , or $\"y%5E2\"$ .\r
\n" ); document.write( "\n" ); document.write( "y^2-24y-256=0\r
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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:
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\n" ); document.write( " $\"y%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca\"$
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\n" ); document.write( " For these solutions to exist, the discriminant $\"b%5E2-4ac\"$ should not be a negative number.
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\n" ); document.write( " First, we need to compute the discriminant $\"b%5E2-4ac\"$ : $\"b%5E2-4ac=%28-24%29%5E2-4%2A1%2A-256=1600\"$ .
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\n" ); document.write( " Discriminant d=1600 is greater than zero. That means that there are two solutions: $\"+x%5B12%5D+=+%28--24%2B-sqrt%28+1600+%29%29%2F2%5Ca\"$ .
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\n" ); document.write( " $\"y%5B1%5D+=+%28-%28-24%29%2Bsqrt%28+1600+%29%29%2F2%5C1+=+32\"$
\n" ); document.write( " $\"y%5B2%5D+=+%28-%28-24%29-sqrt%28+1600+%29%29%2F2%5C1+=+-8\"$
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\n" ); document.write( " Quadratic expression $\"1y%5E2%2B-24y%2B-256\"$ can be factored:
\n" ); document.write( " $\"1y%5E2%2B-24y%2B-256+=+1%28y-32%29%2A%28y--8%29\"$
\n" ); document.write( " Again, the answer is: 32, -8.\n" ); document.write( "Here's your graph:
\n" ); document.write( " $\"graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-24%2Ax%2B-256+%29\"$

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\n" ); document.write( "\n" ); document.write( "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. \n" ); document.write( "

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