1. something i understood:
- there can not be an inverse without an original function
- i have learned that in order to find the inverse of a graph, we switch the input and output of the original graph. here's an example:
- f(ƒ−1(x)) = ƒ−1(f(x)) = x
2. what i do not understand:
many things that i did not understant
- how to graph log.. uh huh. thats right. not at all.
- i also do not under stand how to find x in a case like this: ln (y-1) - ln 2 = x + ln x
(it is when one x is behind "ln" and one is not. i do not know how to separate that x from ln x and put the x's to one side)
THIS TOPIC IS FRIKIN HARD!!!!
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um...didn't that problem call for you to solve for y, not x? if it called for you to solve for y, combine (y-1) and 2 via quotient rule, make everything a power of e to eliminate the ln, then solve for y. if it's just curiosity and you want to know how would you find x in a case like this...yea, i wouldn't know either (well, not at first glance, i'd probably have to mess with the problem for a few minutes to see if i could actually get an answer or not).
ReplyDeleteto graph logs: 1)use a graphing calculator if possible 2)if you cant, take the original exponential function and just draw the same graph reflected across the line of y=x (i could explain the entire process, but im too lazy to type all that, look at some of my other comments on other people's blogs, i explain it on some).
Not really sure how to solve the provided problem , But maybe trying to get rid of Ln of X with a Log Property ?
ReplyDeletehey YOU might not probably know me cuz im 2 period but i think I might be able to help you with the ln (y-1) - ln 2 = x + ln x problem.
ReplyDeleteFirst of all you have to know the RULES of logs cuz they are used in this problem as in ln (y-1) - ln 2 where you put them into a fraction (divide) since you are subtracting. Like this
ln(y-1)/2
And then ,OK ln is the same as log except that ln it's used with base of "e"(2.7182..) and log with a base of 10 or other bases like 2,3,4... but base 10 is the default for log that's why it's in calculators.
so yu raise everything having "e" as a base (so the ln disappears)which is the same as you would do with logs but with the 10 being the base to get rid of log and putting the exponents down.
not sure if this is 100% correct so bare with me...
ReplyDeleteln(y-1)-ln2=(x+lnx)e
(y-1)-ln2= ex+x
+ln2=ex+x+ln2
y-1=ex+x+ln2
+1 +1
y=ex+x+ln2+1
i would of believed we needed to cancel out ln on both sides but apparently thats what the books answer is..
another, this problem does ask for y not for x perhaps knowing this you can understand it a little bit more?
hopes this makes sense...heyyy u were up at 11pm as well!! xD
Rocio and Jesus helped me with #35, which is similar to 36 as well.
ReplyDeleteI saw you needed some help on this problem while reading Rafael's comments.
Do the reciprocal of e ^ -x to the original equation given to you.
Then, multiply both sides by e^x in order to get 1/e^x to become "1". Now you should have e^2x+1 = 3e^x.
Next you subtract 3e^x to both sides and get e^2x-3e^x+1= 0.
Now its in the format of a quadratic equation.
You then have to complete the square
Treat e^x as just x
After completing the square you get
3 +/- (square root) 5 / 2.
"After that you take the natural log of everything to cancel out the e and you have your answer"
hello Wendy!
ReplyDeleteI had trouble with the same topics! I have been reading other people's blogs and it has helped me a lot:] i suggest you do the same! haha
Woah... On Stephanie's (Herrera) do you really go from
ReplyDelete(y-1) - ln 2 = x + ln x
to
ln(y-1)-ln2=(x+lnx)e ?