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Using differential calculus, one can demonstrate that the log function is monotonically increasing. I hesitate to use this as you have posted in the Pre-Calculus forum. At this level, it generally suffices to use the horizontal line test on the graph of the general log function and observe that for any horizontal line, the log functions intersects this line only once.Can we prove that $\log(x)$ is one-to-one then? For all $a,b \in\mathbb{R}^{+}$ that $\log(a) = \log(b) \implies a = b$.
Initially I just wanted an explanation. Now I sort of get it, but I'm tempted to learn a proof. Thanks for the link.Proving the uniqueness of the existence of a logarithm is something you might see in real analysis in college. That is probably overkill for what you need but a proof is sketched here.
How rigorous do you need your proof to be? Are you looking for an intuitive explanation for yourself or a proof?
I posted it in this section because I thought the it would be simpler but I don't mind calculus.Using differential calculus, one can demonstrate that the log function is monotonically increasing. I hesitate to use this as you have posted in the Pre-Calculus forum. At this level, it generally suffices to use the horizontal line test on the graph of the general log function and observe that for any horizontal line, the log functions intersects this line only once.
To keep it as simple as possible, consider:I posted it in this section because I thought the it would be simpler but I don't mind calculus.
Can we prove that $\log(x)$ is one-to-one then? For all $a,b \in\mathbb{R}^{+}$ that $\log(a) = \log(b) \implies a = b$.
Thanks. But that relies on $e^{a} = e^{b} \implies a = b$.Here is another way. Suppose that $e$ is the base, (really it can be a genetic base).
[tex]\begin{align*}\log(a)&=\log(b)\\ e^a &=e^b \\e^{a-b} &=1\\ a-b&=0\\a &= b \end{align*}[/tex].
How do you get $a-b = 0$? I see that $e^{0} = 1$ but that amounts to using we're using what we're proving.Here is another way. Suppose that $e$ is the base, (really it can be a genetic base).
[tex]\begin{align*}\log(a)&=\log(b)\\ e^a &=e^b \\e^{a-b} &=1\\ a-b&=0\\a &= b \end{align*}[/tex].
It's using the fact that given a base, $b>1$ and, and \(\displaystyle b^x=1\), then $x=0$ is a unique solution to this. However, using this fact without proof is not rigorous and needs to be proven. I don't think you're going to see an answer that is truly rigorous without going into some high level math like in the link I gave you.How do you get $a-b = 0$? I see that $e^{0} = 1$ but that amounts to using we're using what we're proving.
How do you get $a-b = 0$? I sey that $e^{0} = 1$ but that amounts to using we're using what we're proving.
That is a very good point.Plato,
Isn't it also necessary to exclude $b=1$ and $b=-1$?