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My algebra textbook says that if the base of the logs are the same in a logarithmic equation, then the one-to-one property can be used to set the arguments equal to each other. My textbook doesn’t provide any explanation why this works or what the one-to-one property is. The only thing about one-to-one functions my textbook has mentioned is that each domain value has a unique range value and vice versa, and if a function is one-to-one, it has an inverse. I’m not sure why these facts allows for the arguments of the log to be equal to each other if the bases are the same. My current understanding of math is at a very basic algebra level, possibly at the level of high school mathematics.

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I'm assuming you mean something like this: $$\log_b(x)=log_b(y)\implies x=y$$ We can say this because $f(x)=\log(x)$ is an injective or one to one function. We say that a function $f$ is injective if $$f(x_1)=f(x_2)\implies x_1=x_2$$ $\log(x)$ maps $x$ to an exponent that raises some base, $b$, to obtain $x$. In other words, if $\log(x)=a$, then $x=b^a$. Suppose that $y\neq x$ but $\log(x)=\log(y)$. Then, $b^a=x\neq y=b^a$, which is a contradiction, so $x=y$ must hold.

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  • $\begingroup$ "Then, if y is some other number" When you say $y$ is some other number, are you saying that it's some other value than $x$? Such as $log (x) = log (y) \Longrightarrow log (2) = log (4)$? $\endgroup$ – Slecker Oct 24 '18 at 0:40
  • $\begingroup$ Or would the one-to-one property mean that $log (x) = log (y) \Longrightarrow log (4) = log (4)$? $\endgroup$ – Slecker Oct 24 '18 at 0:47
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    $\begingroup$ I admit that the wording there was poor, let me edit this. $\endgroup$ – 高田航 Oct 24 '18 at 2:52
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    $\begingroup$ I want to say that $x\neq y$ cannot be true given that $\log(x)=\log(y)$, hopefully this makes more sense now. $\endgroup$ – 高田航 Oct 24 '18 at 2:53
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    $\begingroup$ These are all kind of "mathy" answers and wording. Think of $log_b(x)$ as a simple function on $x$. If two variables $x$ and $y$ produce the same outputs when applied to a function $f$, that is, if $f(x) = f(y)$, and $f$ produces one unique output for every input (that pesky "one to one" business, an example being $f(x) = x+1$), then $x=y$. $\endgroup$ – Russ Oct 24 '18 at 13:10

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