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.
[Math] the one-to-one property and why does it work for logarithms
algebra-precalculuslogarithms
Related Solutions
There are two mistakes in your arguments.
First, the derivatives of $\log_{10}x$ and $\ln x$ are not the same. Since $\log_{b}x=\ln x/\ln b$, we have
$$\frac{\mathrm d}{\mathrm dx}\log_{b}x=\frac{\mathrm d}{\mathrm dx}\frac{\ln x}{\ln b}=\frac1{\ln b}\frac{\mathrm d}{\mathrm dx}\ln x=\frac1{\ln b}\frac1x\;.$$
Second, it's not true that different functions must have different derivatives. Functions that differ only by an additive constant have the same derivative. To use an example close to yours,
$$\frac{\mathrm d}{\mathrm dx}\ln(cx)=c\frac1{cx}=\frac1x\;.$$
This might seem mysterious in that form, but is becomes clearer if you instead write
$$\frac{\mathrm d}{\mathrm dx}\ln(cx)=\frac{\mathrm d}{\mathrm dx}\left(\ln x+\ln c\right)=\frac{\mathrm d}{\mathrm dx}\ln x=\frac1x\;,$$
which shows that this is due to the fact that $\ln (cx)$ and $\ln x$ only differ by an additive constant.
Robert's answer tells you correctly how to find, given an $x$, a complex $c$ such that $x^c=-x$.
However, it is worth pointing out that you cannot find a single $c$ such $x^c=-x$ will hold for every $x$. There's no existing number that achieves this, and if you try to extend the number system by simply postulating that $c$ exists (which I think was the point of the question), you run into this: $$ 0 = \frac08 = \frac{-4+4}{8} = \frac{4^c+4}8 = \frac{2^c2^c+4}8 = \frac{(-2)(-2)+4}8 = \frac{4+4}8 = 1 $$ So if you want your $c$ to exist, but don't want to collapse everything into $0=1$, then you have to discard the exponentiation rule $(ab)^c=a^cb^c$. And without that rule, the result isn't really worthy of being called exponentiation in the first place.
I think you have been fooled by a common introduction to complex numbers, which goes like this: "There is no real number whose square is $-1$, but assume we had such a number anyway and call it $i$. Then we can calculate with the assumption $i^2=-1$, and, ta-dah! we have invented a new kind of numbers". This presentation, however, sweeps an important point under the carpet, namely that we had better be sure that assuming $i^2=-1$ doesn't allow us to prove falsehoods such as $0=1$ about the real numbers that we already know and love -- because if it does that means that we haven't invented a new kind of numbers but simply deceived ourselves.
For $i^2=-1$ it turns out that it's not too difficult to prove that it doesn't lead to nonsense. But even when this proof is shown, something is still swept under the carpet here, namely that if we assume $i^2=-1$ and the familiar rule that every number is either 0 or positive or negative (and the usual rules for positive and negative numbers), then we can still derive nonsense. So what's really true is that we can get a $\sqrt{-1}$ if and only if we're willing to drop what we know about positive and negative numbers (or "less than"/"greater than") when we're working with complex numbers.
Extending the concept of number is almost always connected to a loss of some rules that used to hold before the extension. How far we can push the extension depends on how many rules we're willing to let go of. So you can have your $c$, but the cost you must pay is the rules for how exponentiation works. And if you pay that cost, having $c$ becomes pretty useless anyway, so the whole expedition turns out to be pointless.
(But of course we couldn't have known that before we went and checked. The idea was good enough; it just didn't work. Perhaps your next one will.)
Best Answer
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.