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: 
Consider the function
$$f(x)=x-x^k-8$$ where $k=\log\_{10}(2) <1$ also implies that we work with $x \geq 0$. 

Compute the derivatives
$$f'x)=1-k\, x^{k-1} \qquad \text{and} \qquad f''(x)=-k(k-1)\,x^{k-2}$$

The first derivative cancels only once at
$$x\_\*=\left(\frac{1}{k}\right)^{\frac{1}{k-1}}$$ and, since $k < 1$, by the second derivative test, this point corresponds to a minimum since $f''(x)> \,\, \forall k <1$. For $x >x\_\*$, $f'(x) >0$. Since, we also have $f(0)=-8$, there is only one root to the equation for any $k <1$.

I give you below the solution for various $k$

$$\left(
\begin{array}{cc}
 k & \text{solution} \\
 0.05 & 9.11684 \\
 0.10 & 9.24914 \\
 0.15 & 9.39948 \\
 0.20 & 9.57106 \\
 0.25 & 9.76787 \\
 0.30 & 9.99496 \\
 0.35 & 10.2588 \\
 0.40 & 10.5680 \\
 0.45 & 10.9339 \\
 0.50 & 11.3723 \\
 0.55 & 11.9053 \\
 0.60 & 12.5658 \\
 0.65 & 13.4038 \\
 0.70 & 14.5010 \\
 0.75 & 16.0000 \\
 0.80 & 18.1769 \\
 0.85 & 21.6504 \\
 0.90 & 28.1834 \\
 0.95 & 45.9336
\end{array}
\right)$$
$x =10$ 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.