# How do you write an exponential equation that passes through (1, 1.5), (-1, 6)?

One way the function can be written is as $y = f \left(x\right) = a \cdot {b}^{x}$ and the goal is to find $a$ and $b$. Since $f \left(1\right) = 1.5$ and $f \left(- 1\right) = 6$, we get the following system of two equations and two unknowns: $1.5 = a b$, $6 = \frac{a}{b}$. One of the many ways to solve this system is to divide the second equation by the first to get $4 = \frac{6}{1.5} = \frac{\frac{a}{b}}{a b} = {b}^{- 2}$ so that ${b}^{2} = \setminus \frac{1}{4}$ an $b = \setminus \frac{1}{2}$. This then implies that $a = 6 b = 6 \cdot \frac{1}{2} = 3$ and the answer is $y = 3 \cdot {\left(\frac{1}{2}\right)}^{x} = 3 \cdot {2}^{- x}$.
Another way this type of problem is often solved is to write the function as $y = f \left(x\right) = a \cdot {e}^{k x}$ and the goal is to find $a$ and $k$. The same points as above give the following system of equations: $1.5 = a {e}^{k}$, $6 = a {e}^{- k}$. Dividing the second equation by the first results in $4 = \frac{6}{1.5} = \frac{a {e}^{- k}}{a {e}^{k}} = {e}^{- 2 k}$ so that $- 2 k = \ln \left(4\right)$ and $k = - \frac{1}{2} \cdot \ln \left(4\right) = \ln \left({4}^{- \frac{1}{2}}\right) = \ln \left(0.5\right) \setminus \approx - 0.693147$. Then $a = 6 {e}^{k} = 6 {e}^{\ln \left(0.5\right)} = 6 \cdot 0.5 = 3$ and the answer can be written in approximate form as $y = f \left(x\right) \setminus \approx 3 {e}^{- 0.693147 x}$.