Friday

August 1, 2014

August 1, 2014

Posted by **Kenya** on Sunday, May 1, 2011 at 11:45am.

- Algebra -
**MathMate**, Sunday, May 1, 2011 at 12:00pmLet's try this for f(x)= y = e^(6x)

Step 1:

interchange x and y to get

x=e^(6y)

Step 2:

solve for y in terms of x:

ln(x) = ln(e^(6y)) = 6y

y = ln(x)/6

So

f^{-1}(x) = ln(x)/6

Step 3:

Verify that f(f^{-1}(x))=x (if the inverse was correct).

f(f^{-1}(x))

=f(ln(x)/6)

=e^(6*ln(x)/6)

=e^(ln(x))

=x

So the inverse is correct.

- Algebra -
**Kenya**, Sunday, May 1, 2011 at 12:23pmOk that's slightly confusing.

I understand the flipping of x and y but then what do you do?

so x=log8 y

?

- Algebra -
**MathMate**, Sunday, May 1, 2011 at 1:34pmYes, then you solve for y in terms of x.

use the law of exponents:

e^{log(x)}=x

or

8^{log8 y}=y, etc.

(assuming log8 y is log(y) to the base 8)

raise both sides to power of 8 to get

8^{x}= 8^{log8(y)<?sup> Simplify and solve y in terms of x. } - Algebra - typo -
**MathMate**, Sunday, May 1, 2011 at 1:43pm8

^{x}= 8^{log8(y)}= y

Simplify and solve y in terms of x.

- Algebra -
**Webber**, Sunday, May 1, 2011 at 1:49pmMMMMmmmmmmk that helps, thanks!

- Algebra :) -
**MathMate**, Sunday, May 1, 2011 at 2:54pmYou're welcome!

**Related Questions**

Log Algebra - Find the inverse of the function y=log8 x

Math - Inverse Functions - Find the inverses of the following functions. y = 3(x...

inverse - find f^-1 (x). (this is asking me to find the inverse) f(x) = -(x-2)^2...

Algebra - what is the inverse of the linear parent function? How would you graph...

Algebra 2 - Ok, so I'm trying to find the inverse of each function. We're ...

Pre-Calculus - Find the inverse function for the function f(x)=2^(x)+1/2^(x)-1 ...

graph! - how can the graph of f(x)= X^2-4X be used to obtain the graph of y=g(x...

algebra/trig - 1. f(x)=x^3+5 does f(x) have an inverse? if so, find the inverse ...

Calculus - F(x) = 1-x and g(x) = 1/x These functions have the property that f = ...

algebra 2 - Please help! y=2x I need to find the inverse, and I am wonderingif ...