Simultaneous equation with two unknowns and log 
A mathematical model for a function is $y=\log_ax-b$. If $y=-21.6429$ when $x=105$ and $y=-21.1395$ when $x=211$, find $a$ and $b$ to the nearest integer

Created two equations:
$$-21.6429=\log_a105-b$$
$$-21.1395=\log_a211-b$$
Changed equations to:
$$a^{-21.6429}=105-b$$
$$a^{-21.1395}=211-b$$
Subtracted bottom equation from top:
$$a^{-21.1395}-a^{-21.6429}=106$$
After that I got stuck, and tried to use the calculator solver function, but it gave an out of bounds type error.
 A: You should, right from the start, subtract one equation from the other (here I'll subtract the first from the second). This cancels out $b$:
$$\log_a211-\log_a105=\log_a\frac{211}{105}=-21.1395+21.6429=0.5034$$
$$\frac{211}{105}=a^{0.5034}$$
$$a=\left(\frac{211}{105}\right)^{1/0.5034}=4.00029\ldots\approx4$$
Then we can substitute and derive $b$:
$$b=\log_4{105}+21.6429=24.99984\ldots\approx25$$
The model is $y=\log_4x-25$.
A: Let $k_1=-21.6429, k_2=-21.1395,m_1=105,m_2=211$, then you have this system of equation:
$$k_1=\frac{\log \left(m_2\right)}{\log (a)}-b\\
k_2=\frac{\log \left(m_2\right)}{\log (a)}-b
$$
Which gives us:
$$k_1 \log a-\log m_1=k_2\log a-\log m_2\\
\log a=\frac{\log m_1-\log m_2}{k_1-k_2}\implies a=\exp\left(\frac{\log m_1-\log m_2}{k_1-k_2}\right)$$
Substituting $a$ in either one of the first two equations we get:
$$\begin{align}
b&=\frac{\log m_2-k_2\log a}{\log a}\\
&=\frac{\log m_2-k\left(\frac{\log m_1-\log m_2}{k_1-k_2}\right)}{\frac{\log m_1-\log m_2}{k_1-k_2}}\\
&=\frac{(k_1-k_2)\log m_2-k_2\log m_1}{\log m_1-\log m_2}\\
&=\frac{k_1\log m_2-k_2\log m_1}{\log m_1-\log m_2}
\end{align}$$
Substituting the actual values, you'll get:
$$\{a\to 4.0003,b\to 24.9998\}$$
