How do you find the inverse of 5/e^x+1?

1 Answer
Parabola
Dec 17, 2017

#y=-ln((x-1)/5)#

Explanation:

Remember that #f(x)# is the inverse of #g(x)# if both #f(g(x))# and #g(f(x))# both equal #x#.

From this, we know that when #5/e^x+1=f(x)#, then #5/e^g(x)+1=x# where #g(x)# is the inverse function of #f(x)#.
For simplicity, we will use #y# in place of #g(x)#.

We now isolate #y#.

#5/e^y+1=x#
#5/e^y=x-1#
#5xxe^-y=x-1# We use the fact that #z/x^y=zxx x^-y#
#e^-y=(x-1)/5# We use the natural logarithm on both sides.
#Log _e(e^-y)=Log_e((x-1)/5)# which is really
#ln(e^-y)=ln((x-1)/5)# Remember that #log_x x^y=y#
#-y=ln((x-1)/5)#
#y=-ln((x-1)/5)# That is your answer!

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