# How do you find formulas for the exponential functions satisfying the given conditions f(3)=-3/8 and f(-2)=-12?

##### 1 Answer
Oct 19, 2016

$f \left(x\right) = - 3 {\left(\frac{1}{2}\right)}^{x}$

#### Explanation:

Suppose we have an exponential function of the form $f \left(x\right) = A {e}^{B x}$

Given our conditions, we get the system

$\left\{\begin{matrix}A {e}^{3 B} = - \frac{3}{8} \\ A {e}^{- 2 B} = - 12\end{matrix}\right.$

Multiplying the first equation by ${e}^{- 3 B}$ and the second by ${e}^{2 B}$, we get

$\left\{\begin{matrix}A = - \frac{3}{8} {e}^{- 3 B} \\ A = - 12 {e}^{2 B}\end{matrix}\right.$

Thus, equating the two, we have

$- \frac{3}{8} {e}^{- 3 B} = - 12 {e}^{2 B}$

Multiplying by ${e}^{3 B} / - 12$, we get

$\frac{1}{32} = {e}^{5 B}$

Taking the natural log of both sides, we find

$\ln \left(\frac{1}{32}\right) = \ln \left({e}^{5 B}\right)$

$\implies \ln \left({2}^{- 5}\right) = 5 B$

$\implies - 5 \ln \left(2\right) = 5 B$

$\implies B = - \ln \left(2\right) = \ln \left(\frac{1}{2}\right)$

Note, then, that ${e}^{B x} = {\left({e}^{B}\right)}^{x} = {\left({e}^{\ln \left(\frac{1}{2}\right)}\right)}^{x} = {\left(\frac{1}{2}\right)}^{x}$

Now we can substitute this back into a prior equation, say, $A = - 12 {e}^{2 B}$. Doing so, we get

$A = - 12 {\left(\frac{1}{2}\right)}^{2}$

$= - \frac{12}{4}$

$= - 3$

Thus we get our full equation as

$f \left(x\right) = A {e}^{B x} = - 3 {\left(\frac{1}{2}\right)}^{x}$

Checking we find that

$f \left(3\right) = - 3 {\left(\frac{1}{2}\right)}^{3} = - 3 \left(\frac{1}{8}\right) = - \frac{3}{8}$
and
$f \left(- 2\right) = - 3 {\left(\frac{1}{2}\right)}^{- 2} = - 3 \left(4\right) = - 12$

as desired.