Question

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

Answer 1

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 (1)/(2) - 3 \\ &\approx-3.6931 \end{aligned}`

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 (1)/(2) = e^k`

Take the natural log of both sides:

`\displaystyle \ln (1)/(2) = k\ln e`

Therefore:

`\displaystyle k = \ln (1)/(2)`

Therefore:

`\displaystyle k - 3 = \ln(1)/(2) - 3\approx-3.6931`