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Subtracting a number relates to summing a number, as dividing by a number relates to multiplying by a number, as ??? relates to powering by a number.

May I put it in a more mathematical manner:

$x-y \leftrightarrow x+y\space; \space\space\space\frac{x}{y} \leftrightarrow xy\space ; \space\space\space???\leftrightarrow x^y$

where$\space\space"\leftrightarrow "\space\space$represents a relationship between its sides.

Who is $\space$"???"$\space$ ?

I realise that there are actually two questions here: who is question marks and what is the relation to be considered.

(numbers are over the $\mathbb{R}$ field)

At first it seemed intuitive for me to set $\space\space ???\space\space =\space\space \log_{y}x \space\space\space$;

but then I could find no words to describe the relationship $\space"\leftrightarrow "\space$ which justifies that choice.

Robert Howard
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Leonardo Bohac
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1 Answers1

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All the pairs of operations you mention are called "inverse operations" - you use one to undo the other. For example, if you wanted to solve an equation that involved subtraction, you would use addition: $$x-3=0$$ $$x-3\color{red}{+3}=0\color{red}{+3}$$ $$x=3$$ The same is true for multiplication and division: $$3x=6$$ $$\color{red}{\frac{\color{black}{3x}}3}=\color{red}{\frac{\color{black}6}{3}}$$ $$x=2$$ Now we can turn to exponentiation. Consider the equation $x^y=4$, and let's say we want to solve for $y$. Logarithms are the right way to handle that, so we end up with the following: $$\color{red}{\log_x\color{black}{(x^y)}}=\color{red}{\log_x\color{black}4}$$ $$y=\log_x4$$

Robert Howard
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  • Thanks for clarifying, Robert. I also found a good answer here https://math.stackexchange.com/questions/956776/whats-the-inverse-operation-of-exponents . It turns out that there are two different ways of inverting an exponentiation. – Leonardo Bohac Jul 02 '18 at 21:29
  • Those are certainly some thorough explanations. Glad I could be of some help! – Robert Howard Jul 02 '18 at 21:34