document.write( "Question 1186321: Solve 2^2x+2 - 33(2^x) + 8 = 0. \n" ); document.write( "

Algebra.Com's Answer #817223 by greenestamps(13200) $\"\"$ $\"About$

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\n" ); document.write( " $\"2%5E%282x%2B2%29+-+33%282%5Ex%29+%2B+8+=+0\"$

\n" ); document.write( "Note the powers of 2 in the equation are 2x+2 and x. When you need to solve an equation like this, you need to get the equation in the form of a quadratic by making the larger exponent twice the smaller.

\n" ); document.write( "In this example, the smaller exponent is x, so you want to rewrite the equation with the larger exponent being 2x. That is easy to do:

\n" ); document.write( " $\"2%5E%282x%2B2%29=%282%5E%282x%29%29%282%5E2%29=4%282%5E%282x%29%29\"$

\n" ); document.write( "So rewrite the equation as

\n" ); document.write( " $\"4%282%5E%282x%29%29+-+33%282%5Ex%29+%2B+8+=+0\"$

\n" ); document.write( "Then factor this as a quadratic with $\"2%5Ex\"$ as the variable:

\n" ); document.write( " $\"%284%282%5Ex%29-1%29%282%5Ex-8%29=0\"$
\n" ); document.write( " $\"4%282%5Ex-1%29=0\"$ OR $\"2%5Ex-8=0\"$

\n" ); document.write( "(1) $\"4%282%5Ex-1%29=0\"$
\n" ); document.write( " $\"4%282%5Ex%29=1\"$
\n" ); document.write( " $\"2%5Ex=1%2F4+=+1%2F2%5E2+=+2%5E%28-2%29\"$
\n" ); document.write( " $\"x+=+-2\"$

\n" ); document.write( "(2) $\"2%5Ex-8=0\"$
\n" ); document.write( " $\"2%5Ex=8=2%5E3\"$
\n" ); document.write( " $\"x=3\"$

\n" ); document.write( "ANSWERS: x=-2 or x=3

\n" ); document.write( "CHECK:
\n" ); document.write( "(1) x=-2:

\n" ); document.write( "(2) x=3: $\"2%5E%282x%2B2%29-33%282%5Ex%29%2B8=2%5E8-33%288%29%2B8=256-264%2B8=0\"$

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