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/