Sum of all values that satisfy $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$. 
What is the sum of all values of $x$ that satisfy the equation
  $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$?

I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$
I start with the positive side first:
$$1=\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}$$$$x^2-5x+4=0$$$$x=4, 1$$
Now, I start with the negative side:
$$x=-x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$$$x^\frac{x^2-5x+4}{2x}=-1$$
I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.
$$x^{x^2-5x+4}=1$$
Now, taking the log of both sides, I have 
$$(x^2-5x+4)\cdot\operatorname{log}(x)=0$$
Dividing both sides by $\operatorname{log}(x)$, I get $x^2-5x+4=0$ again, which should give me $x=4, 1$ again. So thus, the answer should be $4+1=5$.
However, this is wrong, as the answer key says that the answer is $-1+1+4=\boxed{4}$. I have checked $-1$ as a solution on all of my equations, and all of them work. How can I derive the $-1$ out from my equations? Or, if my approach or any of my equations are wrong, how do I get the answer $x=-1$?
 A: In your second case, after you render
$x^{\frac{x^2-5x+4}{2x}}=-1$
forget about logarithms.  Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$.  So put $x=-1$ into the equation you derived and see whether, in fact, it's consistent with the power having a value of $-1$.
A: The reasoning here is faulty:

I can't log both sides because of the $-1$, so I raise both sides by a power of $2x$ to get rid of the fraction. Because $2x$ is even, the RHS becomes a $1$.

For one thing, what if $x$ is not an integer, and $2x$ is not even?
Starting from $x^\frac{x^2-5x+4}{2x}=-1$, you could absolute value both sides: $$\lvert x\rvert^\frac{x^2-5x+4}{2x}=1$$
and then take the logarithm: $$\frac{x^2-5x+4}{2x}\ln\lvert x\rvert=0$$ from which you conclude either $x=4$, $x=1$, or $\lvert x\rvert=1$. Check that $4$, $1$, and $-1$ satisfy the original equation, and you have a complete solution.
A: I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$
I start with the positive side first:
$$1=\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}$$$$x^2-5x+4=0$$$$x=\boxed{4}, \boxed{1}$$
Now, I start with the negative side:
$$x=-x^{\frac{1}{2}x-\frac{3}{2}+\frac{4}{2x}}$$$$x^\frac{x^2-5x+4}{2x}=-1$$
Absolute value-ing both sides gives $$\lvert x^\frac{x^2-5x+4}{2x}\rvert=\lvert -1\rvert$$
Now, we use the property that $\lvert x^y\rvert=\lvert x\rvert^y$ to get $$\lvert x\rvert^\frac{x^2-5x+4}{2x}=1$$
Logging both sides gives:
$$\operatorname{log}\lvert x\rvert^\frac{x^2-5x+4}{2x}=\log1$$$$\frac{x^2-5x+4}{2x}\cdot\operatorname{log}\lvert x\rvert=0$$$$\frac{\operatorname{log}\lvert x\rvert\cdot (x^2-5x+4)}{2x}=0$$
Solving for $\operatorname{log}\lvert x\rvert$(multiplying by $2x$ and dividing by $x^2-5x+4$) gives:$$\operatorname{log}\lvert x\rvert\cdot (x^2-5x+4)=0$$$$\operatorname{log}\lvert x\rvert=0$$
Now, solving for $x$ gives:
$$10^{\operatorname{log}\lvert x\rvert}=10^0$$$$\lvert x\rvert=1$$$$x=\boxed{\pm1}$$
So the three solutions are $x=-1, 1, 4$ which gives $-1+1+4=\boxed{4}$
