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### 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/
```
Associated Topics:
High School Basic Algebra
High School Exponents
High School Linear Equations
Middle School Algebra
Middle School Equations
Middle School Exponents

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