Question: I am visualizing what scaling a vector does to its original vector, it can expand it, shrink it, flip it the other way, but is there such thing as a scalar changing the only the $y$ value of a vector?

For instance you have the vector $(2,3)$ does there exist such scalar that would produce $(2,-3)$?

I am just curious. I think there isn't one, but just wanted to check with you guys.

Answer: 
Here is how to "see" the only solution:

* $\lg 2 = \frac{\log\_2 2}{\log\_2 10} = \frac{1}{\log\_2 10}$
* $x^{\lg 2} = 2^{\log\_2 x}$

So, the equation becomes
$$x-8 = x^{\lg(2)} \Leftrightarrow x - 2^{\frac{\log\_2 x}{\log\_2 10}} = 8$$

This equation has obviously the solution $x=10$.

The uniqueness follows (as already mentioned in other solution) from considering the monotonicity of $x - 2^{\frac{\log\_2 x}{\log\_2 10}}$.
$x =10$ is the only solution. Let $f(x) = x - 8 - x^{\lg2}, x\geq0$, then $f'(x)=1-\lg 2 x^{\lg 2 -1}$ and $f''(x) = (\lg2-(\lg 2)^2 )x^{\lg 2 -2}$ which is always non-negative for all $x\geq0$. Hence, $f'(x)$ is an increasing function. 

Suppose $f(x)$ has more than $1$ root, we can denote each root by $x^\*\_i,\ i=1,2,\dots,n$. Note that $x^\*\_{i+1}>x^\*\_i>0$. Then since $f(x^\*\_i)=0$, we must have $x^\*\_i-8={x^\*\_i}^{\lg2}$ for each $i$. Then $f'(x^\*\_i)=1-\lg2+\lg2(\frac{8}{x^\*\_i})>0$. This means that at each root of $f(x)$, the derivative has to be postiive.

Now we know $f(0) = -8 <0$, and $f$ is continuous on $x\geq0$. 

We also know $\lim\_{x\rightarrow0} f'(x)=-\infty$ and $f'(x)$ is continuous on $x>0$, and $f'(x)$ is increasing.

Hence, $f'(x^\*\_1)>0$. For any $x>x^\*\_1$, $f'(x)\geq f'(x^\*\_1)>0$. If $f(x^\*\_2)=0$, by Rolle's thm, $f'(x)=0$ for some $x\in(x^\*\_1,x^\*\_2)$. but this is a contradiction.