document.write( "Question 61660: can somone plese help me solve this?
\n" ); document.write( "log(9, sqrt x ) + log(9, sqrt 2x+3)=1/2
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Algebra.Com's Answer #42484 by psbhowmick(878)\"\" \"About 
You can put this solution on YOUR website!
\"log%289%2Csqrt%28x%29%29++%2B+log%289%2Csqrt%282x%2B3%29%29+=+1%2F2\"
\n" ); document.write( "or \"log%289%2Csqrt%28x%29%2Asqrt%282x%2B3%29%29+=+1%2F2\"
\n" ); document.write( "or \"log%289%2Csqrt%28x%2A%282x%2B3%29%29%29+=+1%2F2\"
\n" ); document.write( "or \"log%289%2C%282x%5E2%2B3x%29%5E%281%2F2%29%29+=+1%2F2\"
\n" ); document.write( "or \"%281%2F2%29log%289%2C%282x%5E2%2B3x%29%29+=+1%2F2\"
\n" ); document.write( "or \"log%289%2C%282x%5E2%2B3x%29%29+=+1\"
\n" ); document.write( "or \"2x%5E2%2B3x+=+9%5E1\"
\n" ); document.write( "or \"2x%5E2+%2B+3x+-+9+=+0\"
\n" ); document.write( "or \"2x%5E2+%2B+6x+-+3x+-+9+=+0\"
\n" ); document.write( "or \"2x%28x+%2B+3%29+-+3%28x+%2B+3%29+=+0\"
\n" ); document.write( "or \"%282x+-+3%29%28x+%2B+3%29+=+0\"
\n" ); document.write( "Either or \"2x+-+3+=+0\" or \"x+%2B+3+=+0\"
\n" ); document.write( "i.e. either \"x+=+3%2F2\" or \"x+=+-3\"\r
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\n" ); document.write( "\n" ); document.write( "But, square root of a negative number is not defined. If \"x+=+-3\" then \"sqrt%28x%29+=+sqrt%28-3%29\" becomes undefined. So x cannot be -3.
\n" ); document.write( "Thus \"x+=+3%2F2\" is the only solution.
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