Review of Exponential and Logrithmic Functions

Exponential Funcions

Using the One-to-one property:

When the bases on both sides of an equation are equal, their powers can be set equal to each other.

Ex 1:

Using the Inverse Property ("The Key to Everything"):

This property allows you to get rid of exponents by taking the log or natural log of both sides of an equation. By doing so, the exponent can be moved to the ouside of the log or natural log, and the log will be able to be changed to 1 (because ln(e)=1 and you can any kind of log so that it changes to 1)

Ex 2:

Logarithmic Functions

Using the One-to-One property:

Ex 3:

Using the Inverse Property ("The Key To Everything"):

Ex 4:

Solving Exponential and Logarithmic Equations- Sec. 3.4

        Friday's lesson consisted of learning the procedures used in solving exponential and logarithmic equations. Before we delved into that, though, we reviewed what it meant for a function to be "one-to-one."

One-to-one:

Where    implies  

for example    3^x=3^2x-5 therefore x=2x-5

and when it comes to logs with the same base implies x=7

Solving Exponential Equations

Example one: Solve for x when

The first thing to recognize is that you can take either an algebraic or logarithmic approach when solving this equation.

If solving this equation algebraically:first take the natural log of both sides
                                                             natural log of e is simply e, so your next step becomes
                                                               x=ln17         which you plug into your calculator to find that
                                                               x=2.83        (remember to round to the nearest hundredth..or else)

If solving this equation logarithmically: first rewrite as the equivalent logarithmic, then get
                                                               x=ln17          (since log base e is representative of ln)
                                                               x=2.83         "  "

Example two: Using log algebraically:  2 + 5^(x-3) = 12  
                                                                 5^x-3=10           subtract 2
                                                           log[5^(x-3)]=log10   take log of both sides
                                                                 (x-3)log5=1        move x-3 out front and log base10 of 10=1
                                                                 x-3=1/log5         divide by log 5
                                                                x=3+(1/log5)       add 3 to each side
                                                                    x=4.43            round to nearest hundredth

                                                                                         (remember that you can easily 
                                                                                               your answer by plugging it
                                                                                                into the original equation)    

The final example Mr.Wilhelm showed us was one of an exponential equation in need of solving that looked eerily similar to a quadratic equation.

Example three:            e^(2x) - 3e^x + 2 = 0         first let u=e^x
                                           u^2 - 3u +2 = 0
                                            (u-2) (u-1) = 0    
                                         e^x = 2    e^x = 1           substitute e^x back in
                                 ln e^x = ln 2     ln e^x = ln 1    take the natural log of all four sides
                                   x = ln 2                 x=0          ln 1 implies e^x must equal 1 so therefore x is 0 and
                                   x= .69                                    

Homework: Section 3.4 problems 3,5,12,14,16-40 multiples of 4, 46, 47, 50

                                                                    
                             

                                                           

                                                               
                                                         

              

Properties of Logrithms

Properties of Logarithms

Assuming and , the Law of everything, which says, , we can conclude that , and Furthermore, we know that and sooo

This is the logic by which the 3, and only 3, properties of logrithms are derived. They are as follows:

The exponential function f with a base a is denoted by
F(x)=a^x
where a>0,a (can not)= 1 , and x is any real number
Domain: (-infinity, infinity) Range:(0, infinity)

f(x)=a*b^(x-c) +d

d= changes vertically (up/down)

C=horizontally (left/Right), changes y intersept

A= vertical stretch or compression

B= Growth rate, larger it is-faster it grows, smaller- slower, less than 1= exponential decay

Compound interest- continually compounded
A=Pe^rt

A= Present Value
P= principal Value
E=2.718(Natural Base)
R=annual interest rate
T=time (years)

Discretely Compounded-
A=p*(1+R/N)^nt

All the same
N=# of compounding periods

a^-x= 1/a^x

a^x *a^y+ a^(x+Y)

a^x/a^y= a^(x-y)

(a^x)^y=a^(x*y)

Logarithms

(the inverse of exponents)

log= common log (log10)

ln= natural log (loge)

the difference between common log and natural log is the base

use logarithms to get x out of the exponent, 5^x = 12

KEY TO EVERYTHING

ex: log101000=3 <---> 10^3=1000

log216=x ---> 2^x=16 ---> x=4

log5125 ---> 5^x=125 ---> x=3

log3(1\18) ---> 3^x=(1\18) ---> x=-4

log42=x ---> 4^x=2 ---> x=(1\2)

Law of Cosines

The Law of Cosines is a method used to solve any triangle, given only a side, an angle, and another side.





It is derived using basic trigonometric identities.

The first step is simplifying c and h:

sinA=

h=c(sinA) b=c(cosA)


Then you use pythagorean theorem on the smaller triangle

a^2=h^2 + (b-x)^2

a^2=[c(sinA)]^2 + [b-c(cosA)]^2

a^2=c^2(sinA)^2 + b^2 - 2bc(cosA) + c^2(cosA)^2

a^2=b^2 + c^2 - 2bc(cosA)

The simplified formula is the standard form of the law of cosines.

There are 3 variations of the standard form of the law of cosines, one for each of the 3 variables involved; a, b, and c.

The only difference between them is which variable is used in the location where a is in the formula above.

The variable that goes there is the one of which you want to know the cosine of. The b and c variables are the other two variables in the triangle.

There is a second form of the Law of Cosines. It is called the alternative form, go figure.

cosA =

There are 3 variations of the alternative form as well, which follow the same rules as the variations for the standard form.


Heron's Formula is a method to find the area of a triangle, knowing just the sides.

Area=

s stands for semi-perimeter, which is, as implied, half of the perimeter;

s=

so the full version of the formula comes out to:

Area =

Dylan Dulberg

Law of Sines

Oblique Triangles - Triangles that have no right angles.
As standard notation, the angles of a triangle are labeled A, B, C, and their opposite sides are labeled a, b, and c.

To solve an oblique triangle, you need to know at least one side and two other parts.
1. Two angles and any side (AAS or ASA) 
2. Two sides and an angle opposite one of them (SSA)
3. Three sides (SSS)
4. Two sides and their included angle (SAS)


Law of Sines

sinA=h/r -> h=b(sinA)         

sinB=h/a -> h=a(sinB)

 =

or

Find X

Area of an oblique triangle

i.e. 

Homework: Section 6.1 #4, 6, 24, 29, 38, 40, 52