# The function f(x)=log base 3 (2^x)-x is injective, domain , range in real numbers?

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You need to test if the given function follows the definition of injective function, such that:

If `f(a) = f(b) => a = b`

Hence, considering `f(x_1) = f(x_2)` , you need to check if `x_1 = x_2` , such that:

`log_3 (2^(x_1)) - x_1 = log_3 (2^(x_2)) - x_2`

Using the property of logarithms yields:

`x_1*log_3 2 - x_1 = x_2*log_3 2 - x_2`

Factoring out `x_1` and `x_2` , yields:

`x_1(log_3 2 - 1) = x_2(log_3 2 - 1)`

Since `log_3 2 - 1 = log_3 2 - log_3 3 != 0` yields:

`x_1(log_3 2 - 1) = x_2(log_3 2 - 1) => x_1 = x_2`

**Hence, considering `f(x_1) = f(x_2)` yields `x_1 = x_2` , thus, the function is injective.**

Graph of function : `f(x)= log_3 (2^x) -x`

`f(x)= log_3 2^x- x`

if `f(x)=f(y)` then:

`log_3 2^x -x=log_3 2^y-y`

`log_3 2^x-log_3 2^y=x-y`

`log_3 2^(x-y)=x-y`

`log_3 2^(x-y)= log_3 2 xx log_2 2^(x-y)=(x-y)log_3 2`

So substituiting in (1):

`(x-y)log_3 2 =(x-y)`

That if `x!=y` `log_3 2=1,`

An abdsurd.

The define set wil be all `RR` for the only problem would be when argument of log (x) is zero, and never will be for `2^x !=0 AA x in RR`

Since range both `2^x, log x, x` are `RR` at the same will be for f(x).