Question

Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)

Answer 1

Answer:

A)

$k=0$

B)

$\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}$

C)

$\displaystyle \begin{aligned} k - 3&= \ln \frac{1}{2} - 3 \\ &\approx-3.6931 \end{aligned}$

Step-by-step explanation:

We are given the function:

$\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}$

A)

Given that h(1) = 20, we want to find k.

h(1) = 20 means that h(x) = 20 when x = 1. Substitute:

$\displaystyle (20) = 20e^{k(1)}$

Simplify:

$1= e^k$

Anything raised to zero (except for zero) is one. Therefore:

$k=0$

B)

Given that h(1) = 40, we want to find 2k + 1.

Likewise, this means that h(x) = 40 when x = 1. Substitute:

$\displaystyle (40) = 20e^{k(1)}$

Simplify:

$\displaystyle 2 = e^{k}$

We can take the natural log of both sides:

$\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}$

By definition, ln(e) = 1. Hence:

$\displaystyle k = \ln 2$

Therefore:

$2k+1 = 2\ln 2+ 1 \approx 2.3863$

C)

Given that h(1) = 10, we want to find k - 3.

Again, this meas that h(x) = 10 when x = 1. Substitute:

$\displaystyle (10) = 20e^{k(1)}$

Simplfy:

$\displaystyle \frac{1}{2} = e^k$

Take the natural log of both sides:

$\displaystyle \ln \frac{1}{2} = k\ln e$

Therefore:

$\displaystyle k = \ln \frac{1}{2}$

Therefore:

$\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931$