How can I show that the exponential function, a x , a ∈ R + − { 1 } is injective? I want to not use the inverse log function as originally I was trying to come up with algebraic justification of why the log function is well-defined and it looks like I can show that assuming the exponential function a x is injective. It looks like showing the function is injective is equivalent to showing that a x =1 implies x=0. Now I am trying to figure out some kind of contradiction if x ≠ 0.

Question

How can I show that the exponential function,       a    x  ,   a  ∈            R        +    −  {  1  } is injective? I want to not use the inverse log function as originally I was trying to come up with algebraic justification of why the log function is well-defined and it looks like I can show that assuming the exponential function       a    x   is injective. It looks like showing the function is injective is equivalent to showing that       a    x  =1 implies x=0. Now I am trying to figure out some kind of contradiction if   x  ≠  0.

Samsonitew7b · Accepted Answer

Well, I guess that your definition of exp involves the ODE      y    ′    =  a  y  ,  y  (  0  )  =  1  →  y  (  t  )  =      exp    a    ⁡  t(this is not   t  →      a    t  , but a does not really matter here).Let us assume that y(x)=1,x&amp;gt;1 (x&amp;lt;1 is similar), and assume that for   t  ∈  [  0  ,  x  ]  ,  y  (  t  )  ≠  1.Then y must be decreasing either near 0 or near x. It is impossible because both derivatives are =a.Then you are done, becausev  exp  :      R    −  {  0  }  ,  +  →      R    ,  ×is a group homomorphism.

inurbandojoa · Accepted Answer

We need to solve the following equation to find y:                              a          y                =        x        →                              (                          e                              ln                ⁡                (                a                )                                      )                                              y              R                        +                          y              I                        i                          =                  e                      ln            ⁡            (            a            )            (                          y              R                        +                          y              I                        i            )                          =        x                  e                      2            π            n            i                              This leads to two equations:                              e                      ln            ⁡            (            a            )                          y              R                                      =        x        →                  y          R                =                              ln            ⁡            (            x            )                                ln            ⁡            (            a            )                                              ln        ⁡        (        a        )                  y          I                =        2        π        n        →                  y          I                =                              2            π            n                                ln            ⁡            (            a            )                                              y        (        n        )        =                              ln            ⁡            (            x            )                                ln            ⁡            (            a            )                          +                              2            π            n                                ln            ⁡            (            a            )                          i            What this means is that, in general, this function is not injective, but if we assume it maps from             R        +    ↦      R  , then there is only a single value of n (n=0) such that this equation is equal (with a real value of y). This means that for each (positive) real value of x there is a single real value, y, such that the equation (or function) holds--this makes the function injective.

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