document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1175406>Question 1175406</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Log0.25 base 8 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #801012 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=801012\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log of .25 to the base of 8 is found as follows:\r<BR>\n" );
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document.write( "use the base conversion formula to convert the log to base 10.\r<BR>\n" );
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document.write( "the conversion formula says that log .25 to the base of 8 is equal to log .25 to the base of 10 divided by log(8) to the base of 10.\r<BR>\n" );
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document.write( "therefore:\r<BR>\n" );
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document.write( "log8(.25) = log(.25)/log(8) using the log function of your calculator because the log function of your calculator is set to base 10, i.e. log(.25) in your calculator is really log10(.25).\r<BR>\n" );
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document.write( "using the calculator, log(.25) to the base of 8 is equal to -2/3.\r<BR>\n" );
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document.write( "the log base conversion formula can also be derived in the following manner.\r<BR>\n" );
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document.write( "start with log8(.25) = y\r<BR>\n" );
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document.write( "this is true if and only if 8^y = .25\r<BR>\n" );
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document.write( "take the log of both sides of this equation to get:\r<BR>\n" );
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document.write( "log(8^y) = log(.25)\r<BR>\n" );
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document.write( "by properties of logarithms, this becomes:\r<BR>\n" );
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document.write( "y * log(8) = log(.25)\r<BR>\n" );
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document.write( "solve for y to get:\r<BR>\n" );
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document.write( "y = log(.25) / log(8).\r<BR>\n" );
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document.write( "your answer was log8(.25) = -2/3.\r<BR>\n" );
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document.write( "by basic definition of logs, this is true if and only if 8^(-2/3) = .25.\r<BR>\n" );
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document.write( "you can use your calculator to verify that 8^(-2/3) = .25.\r<BR>\n" );
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document.write( "note that, by laws of exponents, 8^(-2/3) = 1/8^(2/3).\r<BR>\n" );
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document.write( "you can use your calculator to verify that 1/8^(2/3) = .25.\r<BR>\n" );
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document.write( "all has been confirmed by me through the use of my ti-84 plus calculator.\r<BR>\n" );
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document.write( "your solution is that log8(.25) = -2/3.\r<BR>\n" );
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