document.write( "Question 754773: logx(1/9)=-2
\n" ); document.write( "please help me \n" ); document.write( "

Algebra.Com's Answer #459295 by stanbon(75887) $\"\"$ $\"About$

You can put this solution on YOUR website!
logx(1/9)=-2
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\n" ); document.write( "base is \"x\"
\n" ); document.write( "log or exponent is -2
\n" ); document.write( "result is (1/9)
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\n" ); document.write( "Write in exponential form:
\n" ); document.write( "x^(-2) = 1/9
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\n" ); document.write( "x^(-2) = 3^(-2)
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\n" ); document.write( "x = 3
\n" ); document.write( "===================
\n" ); document.write( "Cheers,
\n" ); document.write( "Stan H. \n" ); document.write( "

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