What is the inverse of # y=log(3x-1)#?

1 Answer
Oct 30, 2015

#y=(log(x)+1)/3#

See the explanation

Explanation:

The objective is to get only #x# on one side of the #=# sign and everything else on the other. Once that is done you change the single #x# to #y# and all the #x's# on the other side of the #=# to #y#.

So first we need to 'extract' the #x# from #log(3x-1)#.

By the way, I assume you mean log to base 10.

Another way of writing the given equation is to write it as:

#10^(3x-1)=y#

Taking logs of both sides

#log(10^(3x-1) )= log(y)#

but #log(10^(3x-1))# may be written as #(3x-1) times log(10)#

and log to base 10 of 10 = 1
That is: #log_10(10) =1#

So no we have

#(3x-1) times 1 = log(y)#

#3x =log(y) +1#

#x= (log(y) +1)/3#

Change the letters round

#y=(log(x)+1)/3#

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