document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=236092>Question 236092</A>: <I><SPAN STYLE=\"word-break: break-all;\"> why the log function graph doesn't go negative?<BR>\n" );
document.write( "what will be the graph of 1/((x^2)-5x+6)? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #173991 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=173991\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> why the log function graph doesn't go negative?\r<BR>\n" );
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document.write( "the log of a number cannot be negative or 0\r<BR>\n" );
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document.write( "this is because the base of the exponential form which is the inverse of the log has to be positive.\r<BR>\n" );
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document.write( "assuming the base of the exponential form has to be positive, it is impossible for the log to be 0 or negative.\r<BR>\n" );
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document.write( "a negative base in the exponential form leads to problems when dealing with exponents that are less than 1.\r<BR>\n" );
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document.write( "-2^(.5) is equivalent to the square root of (-2) which is not a valid numbers so allowing the base to be negative leads to inconsistencies which are avoided byt making the base having to be positive.\r<BR>\n" );
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document.write( "I don't believe you can have a base of 0 for similar reasons.   \r<BR>\n" );
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document.write( "0^0 = 1 because anything to the 0 power = 1\r<BR>\n" );
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document.write( "0^1 = 0\r<BR>\n" );
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document.write( "0^2 = 0*0 = 0\r<BR>\n" );
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document.write( "this also leads to problem that I think are the reasons why the rule is that the base has to be positive.\r<BR>\n" );
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document.write( "assuming the base has to be positive, then the answer to your question becomes:\r<BR>\n" );
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document.write( "the definition of a log is that it is the inverse function of an exponent.\r<BR>\n" );
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document.write( "y = log(b,x) if and only if x = b^y\r<BR>\n" );
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document.write( "the reverse is also true.\r<BR>\n" );
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document.write( "x = b^y if and only if log(b,x) = y\r<BR>\n" );
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document.write( "if you take the equation x = b^y, then you should be able to see that x will never be able to be 0 or negative.\r<BR>\n" );
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document.write( "let's try to make it happen.\r<BR>\n" );
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document.write( "x = b^y is the equation we have to work with.\r<BR>\n" );
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document.write( "if y = 0 then x = b^0 = 1 because anything to the 0 power is equal to 1.\r<BR>\n" );
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document.write( "if y = any positive number greater than 0, then x = b^y becomes a positive number because b has to be positive as stated earlier.\r<BR>\n" );
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document.write( "if -y = any negative number, then by the laws of exponents, x = b^(-y) becomes x = 1/b^y.\r<BR>\n" );
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document.write( "x is still a positive number even though the exponent is negative because of this rule.\r<BR>\n" );
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document.write( "this means that the log of a number can never be 0 or negative because:\r<BR>\n" );
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document.write( "log(b,x) = y if and only if b^y = x and b^y will always be positive assuming that b has to be positive which is the underlying assumption.\r<BR>\n" );
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document.write( "what will be the graph of 1/((x^2)-5x+6)?\r<BR>\n" );
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document.write( "easiest way to find out is to graph it.\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28600%2C600%2C-5%2C10%2C-50%2C50%2C1%2F%28x%5E2-5x%2B6%29%29 ALIGN=MIDDLE  ALT=\"graph%28600%2C600%2C-5%2C10%2C-50%2C50%2C1%2F%28x%5E2-5x%2B6%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "it looks like you have a couple of asymptotes.\r<BR>\n" );
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document.write( "looking at your equation in the denominator of (x^2-5x+6), it appears that it can be factored as:\r<BR>\n" );
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document.write( "(x-2)*x-3)\r<BR>\n" );
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document.write( "multiply these factors out and you get:\r<BR>\n" );
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document.write( "x^2 - 3x - 2x + 6 which becomes:<BR>\n" );
document.write( "x^2 - 5x + 6 which is the same asthe original equation so the factors are good.\r<BR>\n" );
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document.write( "the factors of (x-2) * (x-3) set to 0 yield:\r<BR>\n" );
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document.write( "x = 2 and x = 3\r<BR>\n" );
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document.write( "this means that the denominator of the equation will be 0 when x = 2 and when x = 3.\r<BR>\n" );
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document.write( "this means you have 2 asymptotes at x = 2 and x = 3 as can be seen fairly clearly in the graph.\r<BR>\n" );
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document.write( "draw a vertical line at x = 2 and x = 3 and you will see that the equation will try to reach - infinity and infinity at those points which makes them the asymptotes.\r<BR>\n" );
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document.write( "when x = 1.9999 y equals 9999 + a fraction.<BR>\n" );
document.write( "when x = 2.0001, y equals -10001 - a fraction.\r<BR>\n" );
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document.write( "before the asymptote, the equation is looking to reach infinity the closer it gets to 2.\r<BR>\n" );
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document.write( "after the asymptote, the equation is looking to reach minus infinity the closer it gets to 2.\r<BR>\n" );
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document.write( "same happens around x = 3\r<BR>\n" );
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