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.
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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: $$\{a\to 4.0003,b\to 24.9998\}$$
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 $y=\log_4x-25$.