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 solution since
$$
10^{\lg 2}=2=10-8
$$
To prove that there are no other solutions just plot left and right functions