How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (0,1) and (3,10)?

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How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (0,1) and (3,10)?

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Answer 1 · 2017-03-01T01:00:15

$y = 10^{\frac{x}{3}}$ $y= 10^(x/3)$

Explanation:

We know two $(x, y)$ $(x, y)$ points, so we have enough information to write a system of equations and solve for $a$ $a$ and $b$ $b$ .

Equation 1:

$1 = a e^{0 b}$ $1 = ae^(0b)$

Equation 2:

$10 = a e^{3 b}$ $10 = ae^(3b)$

The first equation can be simplified to $a = 1$ $a = 1$ , because $0 (a)$ $0(a)$ will always equal $0$ $0$ and any real number $x$ $x$ has the property such that $x^{0} = 1$ $x^0 = 1$ .

Solve for $b$ $b$ now.

$10 = 1 (e^{3 b})$ $10 = 1(e^(3b))$

$10 = e^{3 b}$ $10 = e^(3b)$

$ln 10 = ln (e^{3 b})$ $ln10 = ln(e^(3b))$

$ln 10 = 3 b ln e$ $ln10 = 3blne$

$3 b = ln 10$ $3b = ln10$

$b = \frac{1}{3} ln 10$ $b = 1/3ln10$

The function therefore has equation $y = e^{\frac{1}{3} ln 10 x}$ $y = e^(1/3ln10x)$ . Now let's look at simplifying the function.

Use the logarithm property $x^{a} = e^{a ln x}$ $x^a = e^(alnx)$ , to arrive at the result

$y = {(10^{\frac{1}{3}})}^{x} = 10^{\frac{x}{3}}$ $y = (10^(1/3))^x = 10^(x/3)$

Hopefully this helps!

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How do you find the exponential model $y = a e^{b x}$ $y=ae^(bx)$ that goes through the points (0,1) and (3,10)?

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 (0,1) and (3,10)?