document.write( "Question 82737: 2log4+2logx=log32 \n" ); document.write( "
Algebra.Com's Answer #59350 by rapaljer(4671)\"\" \"About 
You can put this solution on YOUR website!
\"2log4%2B2logx=log32\"
\n" ); document.write( "\"log+4%5E2+%2B+log+x%5E2+=+log+32\"
\n" ); document.write( "\"log+16+%2B+log+x%5E2+=+log+32\"
\n" ); document.write( "\"log+16x%5E2+=+log+32+\"\r
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\n" ); document.write( "\n" ); document.write( "\"+16x%5E2+=+32\"
\n" ); document.write( "\"+x%5E2+=+2\"\r
\n" ); document.write( "\n" ); document.write( "\"x=+sqrt%282%29\" \r
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\n" ); document.write( "\n" ); document.write( "(NOTE: \"x=-sqrt%282%29+\" is NOT a valid solution, since you can't have a log of a negative number!)\r
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\n" ); document.write( "\n" ); document.write( "Final answer: \"x=+sqrt%282%29+\"\r
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\n" ); document.write( "\n" ); document.write( "R^2 at SCC\r
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