document.write( "Question 513557: $\"9%5E%28x%2B2%29=240%2B9%5Ex\"$
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Algebra.Com's Answer #343034 by kingme18(98) $\"\"$ $\"About$

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The first thing to recognize is that $\"9%5E%28x%2B2%29=9%5Ex+%2A+9%5E2\"$ by definition of properties of exponents. That is equivalent to saying $\"81%2A9%5Ex\"$ on the left hand side of your equation.\r
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\n" ); document.write( "\n" ); document.write( "You now have $\"81%2A9%5Ex=240%2B9%5Ex\"$ . If you subtract $\"9%5Ex\"$ from both sides, you have $\"81%2A9%5Ex-9%5Ex=240\"$ . The left hand side simplifies to $\"80%2A9%5Ex\"$ (you could factor out $\"9%5Ex\"$ and be left with $\"9%5Ex%2A%2881-1%29=9%5Ex%2A80\"$ or you could recognize that 81*something-same something=80*something).\r
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\n" ); document.write( "\n" ); document.write( "Now, you have $\"80%2A9%5Ex=240\"$ . Divide by 80 on both sides and get $\"9%5Ex=3\"$ . A couple of ways to think about this:
\n" ); document.write( "1) 9 to what power gives me 3?
\n" ); document.write( "2) The square root of 9 is 3; what's the exponent for square root?
\n" ); document.write( "3) $\"log%283%2C9%29\"$ =?
\n" ); document.write( "4) Get like bases. 9 is the same as $\"3%5E2\"$ , so this could be written as $\"%283%5E2%29%5Ex=3%5E1\"$ , or $\"3%5E%282x%29=3%5E1\"$ . Since the bases are equal, the exponents must be equal, and $\"2x=1\"$ , so $\"x=1%2F2\"$ .\r
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\n" ); document.write( "\n" ); document.write( "No matter how you do it, you should get $\"x=1%2F2\"$ . \n" ); document.write( "

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