QuestionDetermine the values of $h$ and $k$ for which the graph of $y=h e^{k x}$ passes through the points $(1,7)$ and $(4,189)$ .
For what values of $h$ and $k$ does the graph pass through the points $(1,7)$ and $(4,189)$ ? $\begin{array}{l} \mathrm{h}=\square \\ \mathrm{k}=\square \end{array}$ (Type exact answers.)

Studdy Solution

STEP 1

1. The function is of the form $y = h e^{kx}$ .
2. The graph passes through the points $(1, 7)$ and $(4, 189)$ .
3. We need to find the values of $h$ and $k$ .

STEP 2

1. Substitute the first point into the equation.
2. Substitute the second point into the equation.
3. Solve the system of equations to find $h$ and $k$ .

STEP 3

Substitute the first point $(1, 7)$ into the equation $y = h e^{kx}$ .
$7 = h e^{k \cdot 1}$ $7 = h e^{k}$

STEP 4

Substitute the second point $(4, 189)$ into the equation $y = h e^{kx}$ .
$189 = h e^{k \cdot 4}$ $189 = h e^{4k}$

STEP 5

Solve the system of equations obtained from STEP_1 and STEP_2.
From STEP_1: $h = \frac{7}{e^{k}}$
Substitute $h$ from STEP_1 into the equation from STEP_2: $189 = \left(\frac{7}{e^{k}}\right) e^{4k}$
Simplify: $189 = 7 e^{3k}$
Divide both sides by 7: $e^{3k} = \frac{189}{7}$ $e^{3k} = 27$
Take the natural logarithm of both sides: $3k = \ln(27)$
Solve for $k$ : $k = \frac{\ln(27)}{3}$ $k = \ln(3)$
Substitute $k = \ln(3)$ back into the expression for $h$ : $h = \frac{7}{e^{\ln(3)}}$ $h = \frac{7}{3}$
The values are: $\begin{array}{l} \mathrm{h}=\frac{7}{3} \\ \mathrm{k}=\ln(3) \end{array}$

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