How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (0,4) and (5, 1/2)?

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How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (0,4) and (5, 1/2)?

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Answer 1 · 2016-07-23T17:52:00

Solve for $a$ $a$ using equation 1 and then for $b$ $b$ using equation 2.

Explanation:

Basically we need to solve for both $a$ $a$ and $b$ $b$ . Since $x = 0$ $x=0$ in the first equation I know that the exponent will be $0$ $0$ and anything raised to $0$ $0$ is evaluates to $1$ $1$ so let's do this first to easily solve for $a$ $a$ .

$4 = a e^{b 0}$ $4=ae^(b0)$
$4 = a \times 1$ $4=a×1$

Having the value of $a$ $a$ we use the second equation to solve for $b$ $b$ .
$\frac{1}{2} = 4 e^{b 5}$ $1/2=4e^(b5)$
$\frac{1}{8} = e^{b 5}$ $1/8=e^(b5)$
$ln (\frac{1}{8}) = b 5$ $ln(1/8)=b5$
$\frac{ln (\frac{1}{8})}{5} = b$ $ln(1/8)/5=b$

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How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (0,4) and (5, 1/2)?

1 Answer

Explanation:

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How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (0,4) and (5, 1/2)?