# If g(x)=2log(x-3)+1 and the base is 10 what is g-1(x)?

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### 2 Answers

We have g(x) = 2 log (x - 3) + 1

Let y = g(x) = 2 log (x - 3) + 1

=> y - 1 = 2 log (x - 3)

=> (y - 1)/2 = log (x - 3)

taking the base of the log as 10

=> 10^[(y - 1)/2] = x - 3

=> x = 10^[(y - 1)/2] + 3

interchange x and y

=> y = 10^[(x - 1)/2] + 3

Therefore the inverse function of g(x) is

**g^-1(x) = y = 10^[(x - 1)/2] + 3**

Given the function g(x) = 2log(x-3)+a.

We are asked to find the inverse function g^-1(x).

Solution:

Let g^-1(x) = y

Then by definition: g(y) = x.

=> 2log(y-3) + 1 = x

=> 2log (y-3) = x-1

=> log (y-3)^2 = log 10^(x-1).

We take the anti logarithms :

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

We take the square root :

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

=> y = 2+10^(x-2)/2.

**So g^-1(x) = 2+10^(x-2)/2 is the inverse function.**