document.write( "Question 332264: Solve: 2^2x - 3*2^x - 40 = 0 \n" ); document.write( "

Algebra.Com's Answer #238285 by jsmallt9(3758) $\"\"$ $\"About$

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$\"2%5E%282x%29+-+3%2A2%5Ex+-+40+=+0\"$
\n" ); document.write( "The key to solving this is to notice that the exponent of 2 in the first term is twice the exponent of the exponent of 2 in the middle term. That makes this equation a quadratic equation in $\"2%5Ex\"$ . If you have trouble seeing this, let's use a temporary variable. Let $\"q+=+2%5Ex\"$ , then $\"q%5E2+=+%282%5Ex%29%5E2+=+2%5E%282x%29\"$ Substituting into the original equation we get:
\n" ); document.write( " $\"q%5E2+-3q+-+40+=+0\"$
\n" ); document.write( "The quadratic nature of the equation should now be clear. We can solve for q by factoring (or using the Quadratic Formula):
\n" ); document.write( " $\"%28q+-+8%29%28q+%2B5%29+=+0\"$
\n" ); document.write( "Using the Zero Product Property:
\n" ); document.write( " $\"q+-+8+=+0\"$ or $\"q+%2B+5+=+0\"$
\n" ); document.write( "Solving we get:
\n" ); document.write( " $\"q+=+8\"$ or $\"q+=+-5\"$
\n" ); document.write( "Now we can substitute back for the temporary variable:
\n" ); document.write( " $\"2%5Ex+=+8\"$ or $\"2%5Ex+=+-5\"$
\n" ); document.write( "From the first equation we can see that x must be 3. And since 2 to any power can never be negative, there is no solution for $\"2%5Ex+=+-5\"$ .

\n" ); document.write( "So the only solution to
\n" ); document.write( " $\"2%5E%282x%29+-+3%2A2%5Ex+-+40+=+0\"$
\n" ); document.write( "is
\n" ); document.write( "x = 3

\n" ); document.write( "With some practice you will no longer need a temporary variable. You will be able to go straight from
\n" ); document.write( " $\"2%5E%282x%29+-+3%2A2%5Ex+-+40+=+0\"$
\n" ); document.write( "to
\n" ); document.write( " $\"%282%5Ex+-+8%29%282%5Ex+%2B+5%29+=+0\"$ \n" ); document.write( "

\n" );