How do you find the inverse of #f(x)=2log(x+1)#?

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Answer 1 · 2015-12-12T17:25:19

I think this is how you write it! #color(white)(...)f^(-1)(x) = 10^(x/2)-1#

Consider log to base 10 of x =y

#log_10(x)=y#

From this we have: #10^y=x#

Now look at your question:

Write as: #y=2log_10(x+1)#

#y/2=log_10(x+1)#

so #10^(y/2)=x+1#

#x=10^(y/2)-1#

Now swap the y's and x's around

#y=10^(x/2)-1#

#f^(-1)(x) = 10^(x/2)-1#