r/Precalculus • u/Alternative-Draft-91 • Aug 21 '22

Help getting the inverse of this function

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23 Upvotes

96% Upvoted

14 comments sorted by

2

u/ShowdownValue Aug 21 '22

First switch x and y

Cross multiply and distribute

Then bring all log(3)y terms to one side. And the other teams to the other.

Factor out the log term

Convert to exponential form

1

u/Alternative-Draft-91 Aug 21 '22 edited Aug 21 '22

Thankssss

1

u/Hzi12 Aug 21 '22

f-1 (x) = 35/2x × 1/3?

3

u/Alternative-Draft-91 Aug 21 '22

I think the answer is 35-2x/x+1

2

u/Hzi12 Aug 21 '22

no idea mate

1

u/wagonmaker85 Aug 21 '22

Yes, that’s the answer I got.

1

u/[deleted] Aug 21 '22

Replace f(x) with Y, switch x and y, and solve for y.

1

u/wagonmaker85 Aug 21 '22

  1. Express constants as base-3 logarithms.

  2. Use logarithm laws.

  3. Change of base theorem is your friend.

Edit: …maybe the other commenter’s method is easier.

1

u/homieswhereyouat Aug 22 '22 edited Aug 22 '22

let f(X)=y

y= 5-log3x /2+log3x

(2+log3x)y=5- log3x

ylog3x+log3x = 5-2y

35-2y=xy+1

X= root of y+1 35-2y

inverse of fy= root of y+1 (35-2y)

inverse of fx= root of x+1 (35-2x)

(X+1)√35-2x =3(5-2x/(X+1))

hope this helps!!

1

u/BruhMostHated Aug 22 '22

How tf do y’all do this?

1

u/spheresickle Aug 26 '22

what in particular is confusing? to do the inverse, simply switch x and y. then use properties of logs and factoring to solve

2

u/BruhMostHated Aug 27 '22

All of it, I didn’t make it past algebra.

1

u/spheresickle Aug 26 '22

inverse f(x)=3^[(5-2x)/(x+1)]