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Mathematics 7 Online
OpenStudy (anonymous):

find the inverse of the function?? y=logbase4 x-2

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)?

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

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?

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 \[x+2=\log_4y\]

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

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

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

OpenStudy (anonymous):

4^x+2 =y

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