document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=921991>Question 921991</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If log 16 64 =2 x + 1, what is the value of x ? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #559307 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=559307\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log16(64) = 2x + 1\r<BR>\n" );
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document.write( "this is true if and only if 16^(2x+1) = 64\r<BR>\n" );
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document.write( "you could solve this by taking the log10 of both sides of the equation and you will find the answer.\r<BR>\n" );
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document.write( "i'll do that after i show you another way, if you recognize it.\r<BR>\n" );
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document.write( "16 = 4^2, so 16^(2x+1) = (4^2)^(2x+1) = 4^(4x+2)\r<BR>\n" );
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document.write( "16^(2x+1) = 64 is equivalent to 4^(4x+2) = 64\r<BR>\n" );
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document.write( "if you know that 64 = 4^3, then you can make your equation become:\r<BR>\n" );
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document.write( "4^(4x+2) = 4^3\r<BR>\n" );
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document.write( "this is true if and only if 4x + 2 = 3<BR>\n" );
document.write( "solve for x to get x = 1/4\r<BR>\n" );
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document.write( "your original equation becomes:\r<BR>\n" );
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document.write( "log16(64) = 2(1/4) + 1) which becomes:\r<BR>\n" );
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document.write( "log16(64) = 3/2.\r<BR>\n" );
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document.write( "this is true if and only if 16^(3/2) = 64\r<BR>\n" );
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document.write( "16^(3/2) is equal to (16^(1/2))^3 which is equal to 4^3 which is equal to 64, so your value of x is good.\r<BR>\n" );
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document.write( "if you did not recognize all of the above, then you could have solved  this problem as follows:\r<BR>\n" );
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document.write( "log16(64) = 2x + 1 if and only if 16^(2x+1) = 64\r<BR>\n" );
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document.write( "take the log10 of both sides of this equation to get:\r<BR>\n" );
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document.write( "log(16^(2x+1) = log(64)\r<BR>\n" );
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document.write( "since log10 is normally just shown as log, there's no loss of accuracy here.\r<BR>\n" );
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document.write( "since log(16^(2x+1) = (2x+1) * log(16), this equation becomes:\r<BR>\n" );
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document.write( "(2x + 1) * log(16) = log(64)\r<BR>\n" );
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document.write( "divide both sides of this equation by log(16) to get:\r<BR>\n" );
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document.write( "2x + 1 = log(64) / log(16)\r<BR>\n" );
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document.write( "use your calculator log function to get:\r<BR>\n" );
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document.write( "2x + 1 = 1.5\r<BR>\n" );
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document.write( "subtract 1 from both sides of this eqaution to get:\r<BR>\n" );
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document.write( "2x = .5\r<BR>\n" );
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document.write( "divide both sides of this eqaution by 2 to get:\r<BR>\n" );
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document.write( "x = .5 / 2 which makes x = 1/4.\r<BR>\n" );
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document.write( "your could also have solved this using the log conversion formula of:\r<BR>\n" );
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document.write( "log16(64) is equivalent to log10(64)/log10(16)\r<BR>\n" );
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document.write( "log10(64)/log10(16) = 2x + 1\r<BR>\n" );
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document.write( "since log10 is normally just shown as log, you get:\r<BR>\n" );
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document.write( "log(64) / log(16) = 2x + 1 which is the same as:\r<BR>\n" );
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document.write( "2x + 1 = log(64) / log(16)\r<BR>\n" );
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document.write( "that's the same equation we got above when we took the log of both sides of the equation, so the answer will be the same.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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