- The inverse of B^x
- All Logarithm functions pass through (1,0) because anything to the power of 0 is 1
2)Inverses
- The equation for Inverses is f(x)= f^-1(x)
- the Output and the input change positions. Y is X and X is Y
- Symmetrical on a graph with the original function on the Y=X line
- Slope is consistent between the symmetrical functions.
- Not all functions have inverses.
Things I do not understand
- When using Ln or Log do you always (exponiate) (spelling error )the previous equation?
- Does Log X always have to be the same on the other side of the equation ?
- What are some things to eliminate before attempting to solve the Log function ?

In order to solve for log then one must eliminate both sides of log by implying either ln or e...okay an example of dealing with ln but im not 100% is right:
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..
I dont belive that he Log of X has to be the same one both sides... in order to cancel it out i think it does.....
ReplyDeleteumm i think other things to eliminated before starting to work with the ln or log is the i guess what you could call the coefficient in that first get rid of that then, you can start the process :)
ReplyDeleteits not the same its just so that u can get x alone or solve for it
ReplyDeleteyou are so smart. i also had problems with things that u didnt get
ReplyDeleteWell you want to eliminate the simple stuff like a "+2" or a "/2" simple stuff like addition and subtraction before the log. Sometimes you need to attack the log first though
ReplyDeleteisn't used to cancel right but first start easy if possible then go to the log
ReplyDelete"Slope is consistent between the symmetrical functions."
ReplyDeleteWhat do you mean here?
And like others said, you want to get the log function with only it's input by itself first before "exponentiating" both sides.
that's not a real word btw. Just making sure.
ReplyDelete