document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1094566>Question 1094566</A>: <I><SPAN STYLE=\"word-break: break-all;\"> could you please help me evaluate this expression <BR>\n" );
document.write( "log(base 3) (12) - log(base3) (28) + log(base3) (63) \r<BR>\n" );
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document.write( "I'm usually pretty good at these types of questions but my answer bing 4.5 was wrong  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #709126 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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/cgi-bin/show-face.mpl?user=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=709126\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You need to apply the change-of-base formula. For your calculations, you can use ln or  log_10 as long as you use the same base for both numerator and denominator. I used ln, but either will give you the same result:<BR>\n" );
document.write( "log_b(x) = (log_d(x))/(log_d(b)<BR>\n" );
document.write( "log_3(12) = (ln(12))/(ln(3)) = 2.261<BR>\n" );
document.write( "log_3(28) = (ln(28))/(ln(3)) = 3.033<BR>\n" );
document.write( "log_3(63) = (ln(63))/(ln(3)) = 3.771<BR>\n" );
document.write( "2.261-3.033+3.771 = 2.999 (the answer is probably 3 because of decimal rounding)<BR>\n" );
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