How do you find the inverse of #f(x)=50,000(0.8)^x#?

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Answer 1 · 2015-12-08T18:34:28

#y=(log(x)-log(50000))/log(0.8)#

write as : #y=50000(0.8)^x#

Taking logs:

#log(y) = log(50000)+log(color(white)(.)(0.8)^xcolor(white)(.))#

But #log(color(white)(.)(0.8)^xcolor(white)(.))# is the same as #xlog(0.8)#

Thus
#x=(log(y)-log(50000))/log(0.8)#

Now swap the x'x and the y's giving:

#y=(log(x)-log(50000))/log(0.8)#