What we understand??? Man, I really hate questions like this...Anyway, on to the topic at hand.
What do I understand about logarithms? Well, lets see. The way I view logarithms is this: in exponential form, the base from the logarithm stays the same. The other two numbers just switch places. That is to say, the number to the left of the equals sign in logarithmic form goes to the right in exponential form and the number all by itself (to the right of the equals sign) in logarithmic form becomes the exponent in exponential form. Essentially, all they do is switch places (it's a little mind trick I developed to remind myself how to do those problems, so if you can't understand it, that's understandable). What else?...Oh, right. In problems involving logarithms, to "undo" (cancel out, get rid of) a log sign, you make everything a power of the logarithmic base. If you have a base that needs to be gotten rid of, you simply make everything as part of a logarithm with a base the same as the exponential base (that gets a little confusing, I don't really now how to phrase that correctly to make more sense). In logarithms involving e, the same applies, only you use e as the base or ln, the natural logarithm (whichever the occasion calls for). Writing this and solving problems involving them made me see problems with exponents and logarithms aren't always as easy as they appear (like the problem in one of the homeworks e^x + e^-x=3, which I found out after some time is just treated like a quadratic equation in the end).
On to inverses (joy...). I understand that to find inverse equations, you just switch the x and y and solve for the new y. I also understand that the graphs of an equation and its inverse are symmetrical about the line y=x, so you can somewhat already know and graph the inverse without ever actually having seen it or graphed it (which is good for lazy people like me: less work!).
There is one thing I don't exactly understand. With some inverse problems, when the original function f(x) is given, they give you some sort of condition (like x>0 or something like that). Now, I understand that the condition is necessary otherwise the inverse won't be one-to-one. However, when I do those problems, I either get something different from the book or something that is in no way the inverse (when I plug it back into the original function, I don't get x as the output). I guess what I'm really trying to as is this: when given a function with a condition, do you have to do something different when finding the inverse or do you go about it normally? Those problems have a tendency to throw me off...
Again, if any of my explanations weren't clear, please let me know.
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According to Ms. Hwang, when given conditions, it may change the answer sometimes. For example,on question 19 on homework C1, you have to make the x that is in the answer negative to satisfy the condition. I had gotten f^-1(x)=2-sq.root of x. In the back of the book, it shows that the answer should have really been f^-1(x)=2-(-sq.root of x) because of the condition. I hope this helps a little, since it helped me when I overheard Ms. Hwang explaining it to a classmate.
ReplyDeleteAbout your question... I think you just "go about it normally" and in the end you fix your equation to fit the condition. That's what I did. (My first answer was not like the one in the back of the book).
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Oh and about your reference to the problem: e^ x + e^ -x = 3, How did you solve it?
okay im having trouble posting my explaination because it says "your HTML cannot be accepted; tag is not allowed:OR"
ReplyDeleteoh nice but it post that ^^^
ReplyDeleteAs Rocio stated you go about it normally.
ReplyDeleteThe condition is to hint of the possibility of the number such as x>0 Here we know that x must be greater than zero portraying that you must alter to input back into the function. Okay consider f(x)=x^2 xgreater than equal to 0 the inverse would be (sqrootx)^2 which cancels into x. So normally the inverse would result in being f-1(x)=sqrx but since it tells us that x is smaller or the same as 0 we need to alter our result...modify it into a negative to be less or equal to zero.
(question answered? i'm delusional at 1 in the morning...)
woooo i got it to post but had to erase some text!!!
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ReplyDeleteMan I thought I had #35 but my answer was different from the back of the book, I think I'm close to solving it though..im going to take your advice about it being like a quadratic formula.
ReplyDeleteI agree with dianna and everyone who said that you just work the problem out and not worry about that untill the end, then you adjust your answer to fit the category that it gives you. Not the greatest explanation i kow......please stand by while i try to explain better.,,.......
ReplyDeletehello, i'm pretty sure that wehn given a condition, you solve the problem as you would if the given condition wasn't even there. At the end, you just make the variable negative or positive. For example, if x> 0, the variable will be positive and if x<0, the variable will be negative. Although i'm sure people have told you this already, i'm simply reinforcing their advice.:]
ReplyDeleteOne of the good questions: When a function has a condition like (x>0) it means that it only works in that condition. If the condition is existent than the graph would work, there are only some points that would work. in the case of (x>0), only the points that are greater than 0 would work. In conclusion, conditions are necessary for some functions to work.
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