document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=701467>Question 701467</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If A=log 2 and B=log 3, write the following in terms of A and B:\r<BR>\n" );
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document.write( "a)log 36\r<BR>\n" );
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document.write( "b)log 54\r<BR>\n" );
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document.write( "c)log (8 sqrt 3)\r<BR>\n" );
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document.write( "d)log (20.25)\r<BR>\n" );
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document.write( "e) log (0.8) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #432433 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=432433\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If A=log 2 and B=log 3, write the following in terms of A and B: <BR>\n" );
document.write( "a)log 36 = log(3^2*2^2) = 2log(3)+2log(2) = 2(B + A) <BR>\n" );
document.write( "b)log 54 = log(27*2) = log(3^3)+log(2) = 3log(3) + log(2) = 3B + A<BR>\n" );
document.write( "c)log (8 sqrt 3) = log(2^3*3^(1/2)) = 3log(2) + (1/2)log(3) = 3A + (1/2)B <BR>\n" );
document.write( "d)log (20.25) = log(81/4) = log(81)-log(4) = 4log(3)-2log(2) = 4B-2A <BR>\n" );
document.write( "e) log (0.8) = log(8/10) = log(8)-log(10) = 3log(2)-1 = 3A-1<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.<BR>\n" );
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document.write( " </SPAN>\n" );
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