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C3: Inverse function of g as ln(bx)

Hi all.

I'm stuck on this question:

g : x -> e^(x+a)

The inverse function of g can be written as
g^(-1) : -> ln(bx), x > 0
where b is a positive constant. Express b in terms of a.

So I've done:
ln(y) = lne^(x+a)
ln(y) = x + a
x = ln(y) - a

but the questions wants it in the form ln(bx). I can't figure out how to put in this form. How do you do this?
(edited 6 years ago)
y = e( x + a )

x = e( y + a )

now take the ln of both sides...
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Avatar for wsteve17
wsteve17
OP
3
Original post by the bear
y = e( x + a )

x = e( y + a )

now take the ln of both sides...


Thanks, but don't you end with
lnx = lne^(y+a)
lnx = y+a
y = lnx - a

the question asks for y=ln(bx), do you put the "lnx" and "-a" together?
Original post by wsteve17
Thanks, but don't you end with
lnx = lne^(y+a)
lnx = y+a
y = lnx - a

the question asks for y=ln(bx), do you put the "lnx" and "-a" together?


a = 1 x a

Now use the fact that 1=ln(e)
To combine your answer into a single logarithm, write a in the form ln k and then apply log rules.
Original post by wsteve17
Thanks, but don't you end with
lnx = lne(y+a)
lnx = y+a
y = lnx - a

the question asks for y=ln(bx), do you put the "lnx" and "-a" together?


yes... you have to write a as ln{e(a)} first
Reply 6
Avatar for wsteve17
wsteve17
OP
3
Original post by the bear
yes... you have to write a as ln{e(a)} first


y = ln(x/(e^a))

Thanks for the help, guys. That seems so obvious now. I'd been looking at it too long to spot it!
Original post by wsteve17
y = ln(x/(ea))

Thanks for the help, guys. That seems so obvious now. I'd been looking at it too long to spot it!


good show !

:congrats:

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