Straight Line on the Graph
Date: 09/10/98 at 22:57:46 From: Elsa Subject: e Let f(x) = ae^(kx) where a>0 and k are constants. Show that the graph of y = ln(f(x)) is a straight line.
Date: 09/13/98 at 19:43:05 From: Doctor Teeple Subject: Re: e Hello Elsa, In order to answer this question, there are some facts about ln and e that you need to use. I'll list them now, so we have something to refer to: Fact 1) ln(a*b) = ln(a) + ln(b) Fact 2) ln(e^x) = x This comes from the fact that the natural logarithm (ln) function is the inverse to e. So, essentially, the ln "undoes" the e^x. Fact 3) e^(xy) = (e^x)^y Fact 4) ln(a^b) = b*ln(a) Now, we need to show that y = ln(f(x)) looks like a straight line. A good place to start is to remember what the equation of a line looks like. In general we know that the function of a straight line looks like: y = mx + b where m is the slope and b is the y-intercept. So we want to show that if we take the natural logarithm of f(x), that is ln(f(x)), it will look like the general equation of a straight line. Using the above facts, we can start to manipulate the ln(f(x)) equation: y = ln(f(x)) = ln(a*e^(kx)) plugging f(x) into the equation = ln(a) + ln(e^(kx)) using fact 1 = . . . Continue using the other facts to get the above equation into the format of the equation of a line. For example, in the next step, what can you do to the ln(e^(kx))? One other hint is that since we know a is a constant, we know that ln(a) is a constant. If that's not clear, please write back. Good luck with continuing your equation manipulations. - Doctor Teeple, The Math Forum Check out our web site! http://mathforum.org/dr.math/
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