## Friday, February 5, 2010

### Exponential Functions

I'm getting feedback that exponential functions are giving you extra trouble. I'd appreciate getting feedback to help know how I can clarify the concepts. Here is a summary of some of the key concepts that I'm wanting you to understand:
• bx is not just a formula "b to the power x" but is a new function, which I'm asking you to call expb(x).
• The properties of exponents like bx+y=bx by and (bx)y=bxy become properties of the exponential functions.
expb(x+y)=expb(x)*expb(y)
expb(xy)= [expb(x)]y

• Logarithms are the inverse functions of exponential functions.
expb(logb(x))=x
logb(expb(x))=x

In formula representation, these are written as follows:
blogb(x)=x
logb(bx)=x
• Whenever you see a formula with an exponential, say b3x-2, you should be able to think in both formula and function modes interchangeably.
b3x-2 = b3x b-2 = (b3)x b-2
b3x-2 = expb(3x-2)
The first mode allows us to recognize that b3x-2t is actually of the form Aqx where A=b-2 and q=b3. The second mode is useful to remind us that we really have a composition when we need to compute a derivative or when dealing with inverse functions.
• There is a special base e that is most important because the corresponding exponential function is its own derivative.
exp'e(x)=expe(x)
d/dx[ex]=ex

The natural exponential is written without a base. The corresponding inverse function is called the natural logarithm, ln(x).
exp(ln(x))=x
ln(exp(x))=x

In formula representation, these are written as follows:
eln(x)=x
ln(ex)=x
The x in these formulas, as always, is a placeholder. So any number or formula could be used in place of x.
• Any exponential can be written in terms of the natural exponential. The key is to use the properties of exponents.
b = eln(b)
bx = [eln(b)]x = eln(b) x
Another way to think of this is using composition of inverse functions: exp(ln( ))
bx = exp(ln(bx))
ln(bx) = x ln(b)
bx = exp(x ln(b)) = ex ln(b)

That is, by replacing bx by eln(b) x, we can write any exponential A bx in the form A ekx where k is the number ln(b).
• Derivatives of exponentials use the basic property that exp'(x) = exp(x), or d/dx[ex] = ex. Usually, this also requires the chain rule: d/dx[eu] = eu u'.
d/dx[e2x] = e2x (2) = 2e2x (u = 2x)
d/dx[e-3x] = e-3x (-3) = -3e-3x (u = -3x)
d/dt[e(t2-5t)] = e(t2-5t) (2t-5) = (2t-5)e(t2-5t) (u = t2-5t)

So this was a moderately long list. Perhaps just re-reading it helped you understand something better. Or perhaps you realize something is still confusing. Please post comments explaining why you are finding problems challenging or perhaps explaining what helped you suddenly understand what you had missed earlier.

Anonymous said...

I am having trouble with the first problem on WebWork. i got the first part of it, but I can't figure out how to get "k". I read the blog and saw that you explained how to get Aq^x, or something similar to that. Could you explain how to convert that into the form you want on WebWork--Ae^(kx)?

Brian Walton said...

Take a look at the second to last bullet. This is the key to switching back and forth between the two representations.

Example: f(x) = 2 * 3^(2x)

Method 1: Replace 3 by e^(ln 3).
f(x) = 2*[ e^(ln 3) ]^(2x)
Using properties of exponents, we get to multiply:
f(x) = 2 * e^[ (ln 3) * (2x) ]

So now if you look in the exponent of e, we have (ln 3) * 2 * x. So k=ln(3) * 2.

Method 2: Apply exp(ln( )) around the entire exponential piece.

f(x) = 2*3^(2x) = 2 * exp( ln[3^(2x)] )
Now use the property that powers move in front of logarithms:
f(x) = 2 * exp( (2x) * ln[3] )

Again, notice that the exponential is x times 2*ln(3). So we again find k=2*ln(3)

Anonymous said...

what would Ae in this problem. Is it 2e?

Brian Walton said...

I assume you meant, "What would A be?" A is the constant multiple of the elementary exponential function. So for my example, it is A=2.

Anonymous said...

i thought i posted this comment saturday afternoon, but i forgot the word verification thing.

for the webwork due tonight (i know its a little late to ask questions but i really want to know) number 9 for me says e^4ln(x).

What happens to the '4'? e^ln(x) just equals x right?

i just dont know what to do with the 4

Brian Walton said...

It is true that e^(ln x) = x because we have inverse functions canceling:

e^(ln x) = exp(ln(x))

See how the inverse functions are exactly next to each other? On the other hand, when there is a 4 as well, this interferes with the inverse functions:

e^(4 ln x) = exp(4 * ln(x))

So the inverse functions are interrupted by the multiplication that occurs prior to composition. You must use one of the identities in order to rewrite the expression so that the inverses are immediately composed with one another.

Hint: 4*ln(x) belongs to an identity about the logarithm. What is it equal to? (Make the substitution.) Similarly, e^(4*u) belongs to an identity about exponentials. What is it equal to? Either identity will allow you to see the inverse functions in direct composition.

Anonymous said...

exp(x) literally does not mean anything to me. I have no idea what that means.

So every time I see exp(2x) or exp(x) etc... nothing happens upstairs. Every time I see that, or something similar, it is like looking at hieroglyphics or arabic and I cannot translate either...

Brian Walton said...

I suppose it is good to have an anonymous comment feature here, or you might not have been willing to ask your question. You should re-read the blog and my supplemental reading notes after this explanation and learn to recognize where the original text was trying to introduce this new idea. (Yes, it is new.)

At the beginning of this blog entry, the first bullet say we are introducing a new function b^x, and I said that "I'm asking you to call [it] exp_b(x)." This is just naming the function.

You should be fairly used to things like, "Let f(x)=x^2". That is, "f" is the function that squares things. It might have been better if we called it "sq", so that "sq(x) = x^2" or perhaps "pow_2(x)=x^2" since we raise x to the power 2. Whatever, we name the function in this example does not change the function, squaring the input.

But for exponential functions, we want to refer to these by name, so we can not use a reusable name like "f". We need a permanent name, like "exp".

When we want to clearly state the base, we use a subscript. So:
exp_2(x) = 2^x
exp_3(x) = 3^x
And if we refer to the natural base e, we do not bother with a subscript:
exp(x) = e^x

So whenever you see e^(2x), the number e was raised to the "2x" power. So this is really the same as exp(2x). On the other hand exp(x+y) is saying to raise e to the (x+y) power: exp(x+y) = e^(x+y).

exp is used when we want to explicitly think in terms of functions as functions. The notation of e raised to a power is used when we want to think about the formula related to the function.

I hope this helps.