MATH 155 SUPPLEMENTAL NOTES 3

EXPONENTIAL FUNCTIONS
THE INVERSE FUNCTION OF y = ln x

Recall that the natural logarithm is also called the logarithm of base e. Also, recall that e is called Euler's constant. Using the definition of a logarithm, log _a y = x if and only if a ^x = y, then the inverse of ln x is e^x.

How do the two functions relate? Well, let us look at the graphs of the two functions together.

y =e ^x is the green curve

y = x is the red line

y = ln x is the purple curve

Looking at this graph, notice that y = e ^x and y = ln x are reflected about the line y = x. This is the pictorial proof of that the two functions are inverses of each other. Now let us talk about the domain and the range of each function.

	DOMAIN	RANGE
y = e ^x	(-¥ , ¥ )	(0, ¥ )
y = ln x	(0, ¥ )	(-¥ ,¥ )

Notice that the domain of y = ln x is the range of y = e ^x, and the range of y = ln x is the domain of y = e ^x.

FACT: ln e ^x = x, for all x.

FACT: e ^{ln x} = x, for x > 0.
SOLVING EQUATIONS INVOLVING ln x AND e^x

EXAMPLE 1:

Solve for y: ln (y + 1) - ln 2 = x + ln x

SOLUTION:

ln (y + 1) - ln 2 - ln x = x

ln (y + 1) - ln (2x) = x

y + 1 = 2x e^x

y = -1 + 2x e^x

EXAMPLE 2:

Solve for t: e ^3t = 2

SOLUTION:

e ^3t = 2 ® ln e^3t = ln 2 ® 3t = ln 2 ® t = (ln 2)/3

EXAMPLE 3:

SOLUTION:

Remember that the laws of exponents still apply for exponential functions. Let us review these laws.

1.	e ^xe ^y = e ^{x + y}
2.	e ^-x = 1/e ^x
3.	(e ^x)/(e ^y) = e ^{x - y}
4.	(e ^x) ^y = e ^xy

EXAMPLE 4:

Solve for t: e ^{x^2} e ^{2x +1} = e^t

SOLUTION:

e ^{x^2} e ^{2x +1} = e^t ® e ^{x^2 + 2x + 1} = e ^t ® ln e ^{x^2 + 2x + 1} = ln e ^t ® x ² + 2x + 1 = t

DERIVATIVE OF e ^x

Let us derive the derivative of e ^x. Consider the function y = e ^x, and let us perform logarithmic differentiation.

y = e ^x ® ln y = ln e ^x ® ln y = x ®

FACT: Given y = e ^x, then y' = e ^x.

EXAMPLE 5:

SOLUTION:

EXAMPLE 6:

SOLUTION:

We will have to do the chain rule here, so let u = 2x, and y = e ^u.

EXAMPLE 7:

SOLUTION:

We will have to use the product rule to find this derivative.

EXAMPLE 8:

SOLUTION:

EXAMPLE 9:

SOLUTION:

First of all, let us simplify the expression.

® y = x - ln (1 + e ^x)

EXAMPLE 10:

SOLUTION:

Using the Fundamental Theorem of Calculus,

THE INTEGRAL OF e ^x

FACT:

EXAMPLE 11:

SOLUTION:

Let u = -4x, then du = -4dx or (-1/4)du = dx.

EXAMPLE 12:

SOLUTION:

Let u = t ², then du = 2t dt or du/ 2 = t dt.

EXAMPLE 13:

SOLUTION:

Let u = tan x, then du = sec ²x dx. When x = 0, then u = 0, and x = p / 4, then u = 1.

EXAMPLE 14:

SOLUTION:

Let u = 1+ e ^r, then du = e ^rdr.

INITIAL VALUE PROBLEMS

Remember that initial value problems are a type of differential equation that has an initial condition present. Solving an initial value problem is a two step process. The first step is to integrate the derivative to find the general solution. The second step is to solve for C by using the initial condition.

EXAMPLE 15:

Solve the initial value problem.

SOLUTION:

Step 1:

Let u = p e ^-t, then du = -p e ^-tdt or -du/ p = e ^-t dt.

Step 2:

FINDING THE AREA BETWEEN TWO CURVES

EXAMPLE 16:

Find the area of the "triangular" region in the first quadrant that is bounded above by the curve y = e ^2x, below by the curve y = e ^x, and on the right by the line x = ln 3.

SOLUTION:

First of all, let us graph the region.

The upper curve for this region is y = e ^2x, and the lower curve for this region is y = e ^x. Remember that the height of a slice is the upper curve minus the lower curve, so here is the integral for the area.

Let u = 2x, then du = 2 dx or du/ 2 = dx.

These last two examples are using the new concepts that we have just learned and applying them to concepts that we learned in calculus I. Work through these examples. If you have any questions or problems, feel free to contact me.

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