document.write( "Question 203483: solve:
\n" ); document.write( "log x^2= (log x)^2
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Algebra.Com's Answer #153514 by RAY100(1637)\"\" \"About 
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log(x^2) = (logx)^2,,,,remember loga^b=b*log a
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\n" ); document.write( "2logx = logx * logx
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\n" ); document.write( "2 logx -logx*logx =0
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\n" ); document.write( "factor
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\n" ); document.write( "If 2a-a^2=0,, we would factor,,,a(2-a)=0
\n" ); document.write( "Then we would set,,a=0,,and then set (2-a) =0,,,to solve for a
\n" ); document.write( "Well, let a = \"log x\",, and repeat the exercise.
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\n" ); document.write( "log x(2-log x) = 0
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\n" ); document.write( "log x = 0,,,,,x=1
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\n" ); document.write( "2-log x =0,,,2= log x,,,,x=100
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\n" ); document.write( "check,,,(x=1),,,log1^2 =(log1)^2,,0=0,,,ok
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\n" ); document.write( "(x=100) log100^2 = (log100)^2,,,,4=4,,,,ok
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