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Example:

$ln(x^2+7x-6) - ln(2x) = 0$

$x^2+7x-6=2x$

$x^2+5x-6=0$

$(x+6)(x-1)=0$

$x_1=-6$, $x_2=1$

Now, if I stopped here, I would only get half a point. That is because $0 < x$ and therefore $x_1=-6$ is invalid. $0 < x$ because you can't take the log of a negative number. Except you can. Okay, you get a complex number, it is hard to compute and above our level, but in this instance the final answer is just “$-6$”, so you dont even have to compute $ln(2cdot(-6))$. So why is the $x_1$ soltion invalid? $1/x$, $x=0$ is invalid, but that is not because it is a difficult problem, it is invalid because it enables contradictions. $ln(-1)=x$ won't lead to contradictions even if you dont know what imagninary numbers are, just leave it like $ln(-1)$ uncomputed. If we then got $e^x$ in a follow up question we could compute the whole thing as $-1$. Kinda like if we didn't have negative numbers, we could still simplify $2-4=x$ to $0-2=x$ and then if we got $3+x=a$ we could solve it like this:

$3+x=a$

$3+(0-2)=a$

$3+0-2=a$

$(3+0)-2=a$

$(3)-2=a$

$1=a$

According to the logic I am arguing this doesn't work because $0-2=y$ is invalid so you can't use $y$ in the first place.

So am I in the right for saying that $x_1=-6$ is a valid answer or am I missing something? Is there anything I stated wrong?

There is a fundamental difference between negative numbers and logarithms of negative numbers, as far as equations are concerned.

A number has a single opposite (or negative): the opposite of $2$ is $-2$.

If you want to consider logarithms of negative numbers, you have to take care of the fact that the operation has not a unique result, but infinitely many.

So, writing $operatorname{Log}$ for the logarithm in the complex numbers, the seemingly innocuous equation $operatorname{Log}x=2$ has infinitely many solutions, namely all complex numbers of the form $e^2+2kpi i$, for $k$ an integer.

Thus it's already a bit confusing to think about what's the meaning of $operatorname{Log}x=operatorname{Log}y$; not to mention that if we interpret $2operatorname{Log}x$ as doubling every possible logarithm of $x$ and $operatorname{Log}x+operatorname{Log}y$ as adding every possible logarithm of $x$ with every possible logarithm of $y$, then
$$
2operatorname{Log}xneoperatorname{Log}x+operatorname{Log}x
$$

Don't use what you don't know about.

To make a different case, consider $S=sum_{n=0}^infty 2^n$. You don't know about series, but this seems to mean $1+2+4+dotsb$, fairly simple. Then
$$
2S=2+4+8+dotsb
$$
Look! If we add $1$ to both sides we get
$$
1+2S=1+2+4+8+dotsb=S
$$
Uh, oh! This means $S=-1$.

The fact that you applying formal properties to objects doesn't mean the properties won't give contradictory results. You have to prove that no contradiction can arise.

By the way, the derivation above of $S=-1$ is perfectly fine if we interpret it in the $2$-adic numbers, but certainly not in the real numbers.

Or you can prove that, in the field of fractions of the ring of formal power series over the rational numbers,
$$
sum_{n=0}^infty x^n=frac{1}{1-x}
$$
but this doesn't mean you can substitute $x=2$ in that equality and “prove” that $S=2$, because substitution is not allowed in that field.

It's a difference of solving the equation over $mathbb{R}$ versus over $mathbb{C}$. If you are asked to solve the equation over $mathbb{R}$, then $x=-6$ is not a solution because the equation is not defined at $x=-6$ in $mathbb{R}$. If you are asked to solve it over $mathbb{C}$ then $x=-6$ is a valid solution because the equation is defined at $x=-6$ in $mathbb{C}$.