Review – Exponential and Logarithmic Equations

Logarithms are used to find the number to which the base must be raised in order to become the value in the logarithm.

This is the method used when solving a logarithm problem:

If log_ax = N, then a^N = x.

I will work through two examples:

Example #1:
log₁₀10,000 = x
10^x = 10,000
x = 4

Example #2:
log₆12x = 2
6² = 12x
36 = 12x
x = 3


Natural logarithms, represented by ln, are logarithms with the base of e.

This is the method used when solving a natural logarithm problem:

If e^y = x, then ln(x) = y.

I will work through an example:

e^x = 30
ln(e^x) = ln(30)
x = ln(30)
x = ~3.4012

Review – Mathematical Induction

Mathematical induction is a method of proving certain formulas. It requires proving the formula for the values of n and n + 1. Here is an example:

Use mathematical induction to prove the formula for every positive integer n.

2 + 4 + 6 + 8 + . . . + 2n = n(n + 1)

First, we must prove this formula true for n = 1:

2(1) = (1)((1) + 1)
2 = 1(2)
2 = 2

Next, we must assume that the formula is true for n, and attempt to prove it for n + 1:

2 + 4 + 6 + 8 + . . . + 2(n + 1) = (n + 1)((n + 1) + 1)
2 + 4 + 6 + 8 + . . . + 2n + 2 = (n + 1)(n + 2)

We can substitute the 2 + 4 + 6 + 8 + . . . with n(n + 1), since we assume that the formula is true for n:

2n + 2n + 2 = (n + 1)(n + 2)
n(n + 1) + 2n + 2 = (n + 1)(n + 2)
n² + n + 2n + 2 = n² + n + 2n + 2
n² + 3n + 2 = n² + 3n + 2

Since the left-hand side now is equal to the right-hand side, we have proven this formula true for every positive integer n.

Mathematical induction can also be used in formulas using summation (sigma) notation.

Review – Parametric Equations

Parametric equations are a method of graphing using an additional parameter, which represents progressive variables such as time. Mostly, this variable is denoted by t. I will provide a quick example to clarify this method of graphing:

x = 2t

y = t²

First, we need to eliminate the variable t. To do this, we need to find a way to make x and y equal to the same value.

x/2 = t
√y = t

Now that t is equal to both x/2 and √y, we can simply set the following:

√y = x/2
y = (1/2)x²


Now, all we have to do is graph this equation:

We could have also graphed this by creating a table by incrementing t and then determining the values of x and y for those values of t.

Since t is not shown on the graph, we must label each point plotted on the line with t = (value of t at that point), so as to indicate the direction in which the graph is progressing. Also, an arrow can be used to indicate this direction.

Review – Gaussian and Gauss-Jordan Elimination

Gaussian Elimination involves reducing a matrix to row-echelon form using the elementary row operations of matrices.
Gauss-Jordan Elimination involves reducing a matrix to reduced row-echelon form using the elementary row operations of matrices.
Either of these methods can be used to solve a system of equations.

Here is an example of solving the following system of equations using the Gauss-Jordan Elimination method:

2x	–	y	+	3z	=	24
		2y	–	z	=	14
7x	–	5y			=	6

Review – Vectors and Dot Products

A vector is a quantity having direction and magnitude.

For any vector u = <a, b>, the magnitude is given by:

||u|| = √(a² + b²)

For any two vectors u = <a, b> and v = <c, d>, the dot product is given by:

u • v = ac + bd

The angle between any two vectors u and v is given by:

cosθ = (u • v) / (||u|| ||v||)

Now, let us try an example of each. For these examples let

u = <6, 4>
and
v = <2, 8>.

First, let us find the magnitude of each vector:

||u|| = √(6² + 4²) =  √(36 + 16)  = √52 ≈ 7.2
||v|| =  √(2² + 8²) =  √(4 + 64) =  √68 ≈ 8.2

Next, let us find the dot product of these vectors:

u • v = (6)(2) + (4)(8) = 12 + 32 = 44

Finally, let us find the angle between these vectors:

cosθ = (u • v) / (||u|| ||v||)
cosθ ≈ 44 / (7.2 × 8.2)
cosθ ≈ 44 / 59
cosθ ≈ 0.75
θ ≈ arccos(0.75)
θ ≈ 41.4º

Review – Verifying Trigonometric Identities

To verify a trigonometric identity, you need to use existing trigonometric identities to gradually modify one side of the equation and eventually make it look exactly the same as the other side.

Some tips for verifying identities are the following:

• Try to express everything in sin's and cos's.
• Only work with one side of the equation.

Here is an example of a verifying problem; I will list out the steps as I solve it:

cos(–θ) / [1+sin(–θ)] = secθ + tanθ

Use the Even and Odd identities to simplify the terms containing –θ:

cosθ / (1 – sinθ) =

Multiply the numerator and denominator with the conjugate of the denominator:

[cosθ × (1 + sinθ)] / [(1 – sinθ) × (1 + sinθ)] =

Distribute:

(cosθ + cosθsinθ) / (1 – sin²θ) =

Use the Pythagorean identity sin²θ + cos²θ = 1 to change the denominator to cos²θ:

(cosθ + cosθsinθ) / cos²θ =

Split this fraction into two fractions:

(cosθ / cos²θ) + (cosθsinθ / cos²θ) =

Cancel out the cos's:

(1 / cosθ) + (sinθ / cosθ) =

Use the Reciprocal identity 1 / cosθ = secθ and the Quotient identity sinθ / cosθ = tanθ to rewrite the fractions:

secθ + tanθ = secθ + tanθ

Since the LHS (left-hand side) is now equal to the RHS (right-hand side), the verification is done!

12.2 - Techniques for Evaluating Limits

In the last section, we learned how to solve basic limit problems. However, not all limit problems can be solved this easily. To solve these, we must learn other techniques for evaluating limits.

The easiest method of evaluating a limit is to use the technique of direct substitution. Simply substitute the value which x is approaching into the function after the limit and solve (this is the same example from the previous blog post, so you can visit that for a more detailed walkthrough):

Substitute 2 into the equation to find the answer:

3(2) – 2 = 6 – 2 = 4

Although, sometimes direct substitution will not work because it will result in indeterminate form, or zero over zero. When this happens, you must use either the cancellation technique or the rationalizing technique.

The cancellation technique involves factoring the numerator and denominator, canceling out common factors, and performing direct substitution again. Here is Example 3 from page 878 of our textbook, which demonstrates the cancellation technique:

Use the rationalizing technique when the numerator or denominator involve roots. With this method, you multiply both the numerator and denominator by the conjugate and perform direct substitution again. Here is Example 4 from page 879 of our textbook, which demonstrates the rationalizing technique: