Saturday, November 14, 2009

Logs and Inverses

I am posting this at this hour because I had two test, one an essay and the other a 20 question quiz, and a discussion for AP online literature...yes online...fairly, the due date is 11:59 pm which lead me to spare 5 minutes before everything was due...Just thought you should know if I tend to mispell or drift off from the subject but i'll try!!...Okay moving on!

Recap 4 major concepts that you have understood about either/both of these two topics.

1) When each output of a value of a function assists with ONE input value which leads to the function of one-to-one. Introducing the Horizontal line test is a great way to see if your graph is a one-to-one function. If this line intersects at one point then it is indeed a one-to-one function, if it intersects at more than once at the y-value then it is not. Basically the value of the range of the function should correspond with only one value in the domain. For instance the graph of y=x^3 meets the one-to-one function as opposed to y=x^2 which is a parabola meaning that using the horizontal test will intersect more that one point(two points)

2) The values of x and y switch to create the inverse of a function f=f-1. The inverse of a function is merely a mirror effect of is folded then an overlap of the objective. Finding the inverse of a function consist of whether the horizontal line intersects the graph at only one place therefore having a one-to-one and an inverse. To solve for x for y:
y= -2x+4
x=-1/2y +2
switch
y=-1/2x+2
To check:
input the inverse into the function or the function into the inverse to receive x.
if x is the result then its the correct inverse.
Graphing both the inverse and original function leads to reflections of one another across the line y=x which sets off as the divider.

3)Logarithmic Functions have inverses as well. The base of a logarithm function y=logax has the inverse of y=a^x
The thing with logarithms if that you may go back and forth such as logex= ln x and log10x= log x. What is considered to be a natural logarithm function is y=ln x and the common logarithm function is y=log x.

4) I understand the properties of logarithms fairly well. The product rule which consists of a result of addition: logaxy= logax + logay, the division rule which leads to the result of subtraction: logx/y= logx - logy, and the power rule with brings the power to the front of the equation: logx^y= ylogx.

Write about what you did NOT understand completely.


Well it seems that everyone else has a problem in graphing logarithm function, i was rather surprised to find myself in that position as well. How is it that we graph? When dealing with the function f(x)=log2x...
okay change the base formula...f(x)=log2x=lnx/ln2
Then do we find ln of x and 3 and input it and divide the number then graph? Or do we input the ln and number into the calculator to result in the correct number because sloving for the number is the algebraical format? (does this make sense?)
Okay i also did not understand some logarithms that were reviewed in class..
log6 1/536^2= log6 1/36^1/5= log6 36^-1/5= log6 (6^2)^-1/5 = 2^-1/5



5 comments:

  1. I just learned how to graph from Jesus, you should check out his explanation on my blog its really helpful.
    ----------------------------------------------
    Oh and I'm going to try to explain to you that logarithmic problem to the best of my ability but I'm going to post it on my blog so that you can understand it better because the comment box is really restricted.

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  2. i am completely lost when it comes to graphing log. but if you take a look back at "intro o logarithm" classwork... we will see that its simply just an inverse function of an exponential function..

    :)

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  3. Although i cannot help you graph logarithmic functions, i can help you better understand how to solve logarithms such as log6 36^-1/5.
    Well first, i would change the 36 to 6^2
    Then, i would just multiply the powers (2 X -1/5).
    o then, the answer would be -2/5.
    The class exercise we did on Tuesday also helps a lot, try looking back on it. Hope this helps!

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  4. hmm doesnt it cancel when log6 (6^2)^-1/5 i mean the 6's cancel giving you (1^2)^-1/5 then just multiply the exponents giving you -2/5 i think thats right...right?

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