# SOLUTION: why the log function graph doesn't go negative? what will be the graph of 1/((x^2)-5x+6)?

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 Click here to see ALL problems on Graphs Question 236092: why the log function graph doesn't go negative? what will be the graph of 1/((x^2)-5x+6)? Answer by Theo(5548)   (Show Source): You can put this solution on YOUR website! why the log function graph doesn't go negative? the log of a number cannot be negative or 0 this is because the base of the exponential form which is the inverse of the log has to be positive. assuming the base of the exponential form has to be positive, it is impossible for the log to be 0 or negative. a negative base in the exponential form leads to problems when dealing with exponents that are less than 1. -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. I don't believe you can have a base of 0 for similar reasons. 0^0 = 1 because anything to the 0 power = 1 0^1 = 0 0^2 = 0*0 = 0 this also leads to problem that I think are the reasons why the rule is that the base has to be positive. assuming the base has to be positive, then the answer to your question becomes: the definition of a log is that it is the inverse function of an exponent. y = log(b,x) if and only if x = b^y the reverse is also true. x = b^y if and only if log(b,x) = y 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. let's try to make it happen. x = b^y is the equation we have to work with. if y = 0 then x = b^0 = 1 because anything to the 0 power is equal to 1. 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. if -y = any negative number, then by the laws of exponents, x = b^(-y) becomes x = 1/b^y. x is still a positive number even though the exponent is negative because of this rule. this means that the log of a number can never be 0 or negative because: 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. what will be the graph of 1/((x^2)-5x+6)? easiest way to find out is to graph it. it looks like you have a couple of asymptotes. looking at your equation in the denominator of (x^2-5x+6), it appears that it can be factored as: (x-2)*x-3) multiply these factors out and you get: x^2 - 3x - 2x + 6 which becomes: x^2 - 5x + 6 which is the same asthe original equation so the factors are good. the factors of (x-2) * (x-3) set to 0 yield: x = 2 and x = 3 this means that the denominator of the equation will be 0 when x = 2 and when x = 3. this means you have 2 asymptotes at x = 2 and x = 3 as can be seen fairly clearly in the graph. 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. when x = 1.9999 y equals 9999 + a fraction. when x = 2.0001, y equals -10001 - a fraction. before the asymptote, the equation is looking to reach infinity the closer it gets to 2. after the asymptote, the equation is looking to reach minus infinity the closer it gets to 2. same happens around x = 3