document.write( "Question 841736: (2^(x+1))=(2^(x-1))+48 \n" ); document.write( "

Algebra.Com's Answer #507179 by josgarithmetic(39630) $\"\"$ $\"About$

You can put this solution on YOUR website!
A rule about exponents: $\"b%5E%28a%2Bc%29=%28b%5Ea%29%2A%28b%5Ec%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "See that in your equation you have a term $\"2%5E%28x%2B1%29\"$ , and you can do this:
\n" ); document.write( " $\"2%5E%28x%2B1%29=%282%5E%28x-1%29%29%282%5E2%29\"$ .\r
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\n" ); document.write( "\n" ); document.write( "Some steps........
\n" ); document.write( " $\"%282%5E%28x%2B1%29%29=%282%5E%28x-1%29%29%2B48\"$
\n" ); document.write( " $\"2%5E%28x%2B1%29-2%5E%28x-1%29=48\"$
\n" ); document.write( " $\"%282%5E%28x-1%29%29%282%5E2-1%29=48\"$
\n" ); document.write( " $\"%282%5E%28x-1%29%29%284-1%29=48\"$
\n" ); document.write( " $\"%282%5E%28x-1%29%29%2A3=48\"$ and then factor 48
\n" ); document.write( " $\"3%2A2%5E%28x-1%29=3%2A2%5E4\"$
\n" ); document.write( " $\"2%5E%28x-1%29=2%5E4\"$ , the bases are the same, so the exponents must be equal.
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\n" ); document.write( " $\"x-1=4\"$
\n" ); document.write( " $\"x=4%2B1\"$
\n" ); document.write( " $\"highlight%28x=5%29\"$ \n" ); document.write( "

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