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OpenStudy (anonymous):
@geerky42
OpenStudy (xapproachesinfinity):
is that \[y=\log_4(x-2) ~~or~~y=\log_4x-2\]
OpenStudy (anonymous):
it has no parenthesis, but there is a space between the 4 and x
OpenStudy (xapproachesinfinity):
so it is the later.
since log is one-to-one we don't have any problem
we can go directly into finding the inverse function
OpenStudy (anonymous):
how do we know if one to one? if we do f(g(x)) and (g(f(x)?
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OpenStudy (xapproachesinfinity):
pretty good question
one to one means any element of domain maps to exactly one element of the range and the other way around
OpenStudy (xapproachesinfinity):
so if we take y=x^2
and try for example (-2) and 2 we see that both of them are giving 4
that means it is not one to one
geerky42 (geerky42):
Another way to tell if the function is one to one is by doing horizontal line test.
OpenStudy (xapproachesinfinity):
to speak clearly i should say one to one correspondence
OpenStudy (xapproachesinfinity):
yeah geometrically we can test this by a horizontal line like @geerky42 mentioned
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OpenStudy (anonymous):
if it passes horizontal test, it's one to one?
OpenStudy (xapproachesinfinity):
at any rate this not something new to you since you are dealing with inverse
the first and most important aspect to look at is this
OpenStudy (xapproachesinfinity):
yes!
OpenStudy (xapproachesinfinity):
that is what you were thought no?
OpenStudy (anonymous):
i know that the logarithm of an exponental function is the inverse ? but how do i know this inverse of exponential?
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OpenStudy (xapproachesinfinity):
well that is want we are going to find out!
and true log are inverse relation with exponential so
this one should give us some exp function of some sort
OpenStudy (xapproachesinfinity):
let's find out
OpenStudy (xapproachesinfinity):
first you switch the y and x roles
\[x=\log_4y-2\]
the game is to solve for y
so you need to have experience on solving such equation
OpenStudy (xapproachesinfinity):
how would you solve for y here?
OpenStudy (xapproachesinfinity):
well let me make it easier
\[x+2=\log_4y\]
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OpenStudy (xapproachesinfinity):
any rule that come in mind to get rid of that log_4
OpenStudy (xapproachesinfinity):
meant "log base 4"
OpenStudy (xapproachesinfinity):
where are you leaving me behind ?
OpenStudy (anonymous):
use change of base?
OpenStudy (anonymous):
@xapproachesinfinity
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OpenStudy (xapproachesinfinity):
nope
OpenStudy (xapproachesinfinity):
anything else?
OpenStudy (xapproachesinfinity):
go and check your notes lol
OpenStudy (anonymous):
i really dont know??
OpenStudy (anonymous):
it's equal to 4^x=y
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OpenStudy (xapproachesinfinity):
how?
OpenStudy (anonymous):
exponential form
OpenStudy (xapproachesinfinity):
where did you got that
OpenStudy (xapproachesinfinity):
well let me take you one step and you do the rest
\[x+2=\log_4y \Longrightarrow 4^{x+2}=4^{\log_4y}\]
OpenStudy (anonymous):
oh
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