document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=220516>Question 220516</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can anyone help me?\r<BR>\n" );
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document.write( "Solve for x:\r<BR>\n" );
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document.write( "log(subscript 3) x^9 = -3 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #165771 by </B><A HREF=/tutors/aboutme.mpl?userid=likaaka><B>likaaka(51)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=likaaka><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=165771\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> This is the basic logarithmic relationship y=b^x is equivalent to log(subscript b)(y)=x<BR>\n" );
document.write( "So in your problem log(subscript 3) x^9 = -3 is equivalent to x^9=3^(-3)<BR>\n" );
document.write( "To solve for x, first get rid of the negative exponent<BR>\n" );
document.write( "x^9 = 1/(3^3), simplify<BR>\n" );
document.write( "x^9 = 1/27, now take the 9th root of both sides<BR>\n" );
document.write( "root9 of x^9 = root9 of (1/27), simplify **sorry couldn't figure out the html code<BR>\n" );
document.write( "x = 1/(root 9 of 27), now since you shouldn't have radicals in the denominator you must rationalize the denominator<BR>\n" );
document.write( "There are three 3s in 27 ie 3*3*3=27, so to rationalize the denominator, we need six more 3s to get rid of the radical. Multiply top and bottom by the 9th root of 3*3*3*3*3*3 or 3^6<BR>\n" );
document.write( "1/(9th root of 3^3) * (9th root of 3^6)/(9th root of 3^6) = <BR>\n" );
document.write( "(9th root of 3^6)/(9th root of 3^9), simplfy<BR>\n" );
document.write( "(9th root of 3^6)/3 </SPAN>\n" );
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