document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=448377>Question 448377</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve for x:<BR>\n" );
document.write( "2^logx= 1/4 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #308533 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=308533\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If you have a problem in which a variable appears in an exponent, one of the things to consider is taking a logarithm of both sides. This problem is:\r<BR>\n" );
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document.write( "2^(logx) = 1/4\r<BR>\n" );
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document.write( "Since the variable x is in the exponent, take the log of both sides:\r<BR>\n" );
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document.write( "log[2^(logx)] = log(1/4)\r<BR>\n" );
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document.write( "One of the properties of logarithms is that exponents come out as multipliers. This rule applies to the left side of this equation. Converting the exponent (log 2x) to a multiplier makes the equation become:\r<BR>\n" );
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document.write( "log(2)*logx = log(1/4)\r<BR>\n" );
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document.write( "Note that log(2) and log(1/4) are just numbers that can be found using a calculator.  Log(2) = 0.301029995 and log(1/4) = log(0.25) = -0.602059991\r<BR>\n" );
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document.write( "Substituting these for log(2) and log(1/4) makes the equation become:\r<BR>\n" );
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document.write( "(0.301029995)*logx = -0.602059991\r<BR>\n" );
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document.write( "Divide both sides of this equation by 0.301029995 and the equation reduces to:\r<BR>\n" );
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document.write( "logx = -2\r<BR>\n" );
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document.write( "Ten is the base of the term \"log\".  By definition \"if log to the base 10 of x equals y this is equivalent to saying the base 10 raised to the exponent y equals x.\" Think of this and become familiar with this form of conversion.\r<BR>\n" );
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document.write( "As this definition applies to the equation:\r<BR>\n" );
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document.write( "logx = -2\r<BR>\n" );
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document.write( "We can say that 10 (the base) raised to the exponent -2 equals x.  In equation form this is:\r<BR>\n" );
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document.write( "x = 10^(-2)\r<BR>\n" );
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document.write( "But by the rules of exponents 10^(-2) = 1/(10^2) = 1/100 = 0.01\r<BR>\n" );
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document.write( "This results in the answer:\r<BR>\n" );
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document.write( "x = 0.01\r<BR>\n" );
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document.write( "You can check this by substituting 0.01 for x in the original problem:\r<BR>\n" );
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document.write( "2^(log(x)) \r<BR>\n" );
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document.write( "Substituting 0.01 for x:\r<BR>\n" );
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document.write( "2^(log(0.01))\r<BR>\n" );
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document.write( "Use a calculator to find that the log(0.01) = -2\r<BR>\n" );
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document.write( "So our term above becomes:\r<BR>\n" );
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document.write( "2^(-2)\r<BR>\n" );
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document.write( "By the rules of exponents, a negative exponent is equivalent to 1 over the term with a positive exponent.  So this becomes:\r<BR>\n" );
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document.write( "2^(-2) = 1/[2^2) = 1/4\r<BR>\n" );
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document.write( "and this is exactly as the original problem statement said it should be. \r<BR>\n" );
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document.write( "When x equals 0.01, the term 2^logx does equal 1/4 or 0.25\r<BR>\n" );
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document.write( "Hope this helps you to understand the problem and how to think your way through it.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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