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84 votes
Accepted

Why are there two inverses to exponentiation?

The concept of an inverse operation itself is a bit tricky. Often we consider arithmetic operations to be binary operations: $\DeclareMathOperator{\add}{add}$ $\DeclareMathOperator{\subtract}{subtract}...
Justin Hancock's user avatar
33 votes

Why are there two inverses to exponentiation?

Is your variable at the base or the exponent? An exponential function is a function of the form $f(x)=a^x$ for some constant $a$. In this case, the inverse is indeed given by a logarithm $f^{-1}(x)=\...
Luiz Cordeiro's user avatar
23 votes

Why are there two inverses to exponentiation?

(Despite good answers already, I thought the concrete example below could be useful.) For multiplication and addition, there is exactly one inverse operation, namely division and subtraction. Yet ...
jpa's user avatar
  • 399
4 votes

Why are there two inverses to exponentiation?

The other answers have shown that it depends on which of the two inputs you fix and which you let vary. As a way to visualize this, here's a plot of the 3D surface $z = x^y$. I've drawn two polynomial ...
JounceCracklePop's user avatar

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