How do you find the inverse of #f(x) =e^(2x-1)#?

1 Answer
Lotusbluete
Nov 16, 2015

#f^(-1)(x) = 1/2(ln x + 1)# for #x > 0#

Explanation:

First, set plug #y# for #f(x)#:

#y = e^(2x-1)#

Then, exchange #y# and #x# in the equation:

#x = e^(2y-1)#

Now, try to solve the equation for #y# in terms of #x#.

To do so, we need to apply the logarithmic function #ln# to both sides of the equation. Please note that #x > 0# needs to hold so that the logarithm is well defined.

#ln (x) = ln(e^(2y-1))#
#<=> ln(x) = 2y-1#
#<=> ln(x) + 1 = 2y#
#<=> 1/2(ln(x) + 1) = y#

The only thing left to do is replacing #y# with #f^(-1)(x)#.
The inverse function is

#f^(-1)(x) = 1/2(ln x + 1)# for #x > 0#

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