# how do i find the inverse of f(x) = 3x + 2

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You need to replace `y` for `f(x),` such that:

`y = 3^x + 2`

You need to solve for x the equation, hence, you need to isolate the term that contains x to one side, such that:

`y - 2 = 3^x `

You need to take common logarithms both side to use the logarithmic identity `log_a b^x = c => x*log_a b = c` , such that:

`ln(y - 2) = ln(3^x) => x*ln 3 = ln(y - 2) => x = (ln(y - 2))/ln 3`

Replace back in the last equation y for x, such that and x for y, such that:

`y = (ln(x - 2))/ln 3`

Change `y` with `f^(-1)(x)` , such that:

`f^(-1)(x) = (ln(x - 2))/ln 3`

**Hence, evaluating the function inverse of f(x) = 3^x+2 , yields **`f^(-1)(x) = (ln(x - 2))/ln 3.`

Original function: `f(x) =3x + 2`

In order to find function INVERSE, we'll say `y = 3x + 2.`

Now solve your equation for x, so we'll first subtract 2 then divide by 3, to get:

`(y-2)/3 = x` Once I have "x =" I can switch the x and y, the "y=" will be the inverse:

`y = ( x-2 ) / 3` **is the inverse.**

its 3 to the x+2 3^x+2 how would i solve that