3.3 The special number e

a. We want to find the input that would give [latex]2[/latex] as an output. We first start by drawing a graph of [latex]f(t) = e^t[/latex] in desmos and the horizontal line [latex]y = 2[/latex].

Using the graph this gives an input of about [latex]0.693[/latex]. So [latex]t \approx 0.693.[/latex]

b. We want to find the input that would give [latex]5[/latex] as an output. We start by drawing a graph of [latex]f(t) = e^t[/latex] in desmos and the horizontal line [latex]y = 5[/latex].

Based off the graph of [latex]e^t[/latex], this input is about [latex]1.609[/latex]. In our case our input is [latex]3t[/latex] so we have that [latex]3t = 1.609[/latex]. Dividing both sides by [latex]3[/latex] gives [latex]t \approx 0.563[/latex].

c. We will first do some algebra to get [latex]e^t[/latex] by itself on one side so we can use the graph to find a solution:

\begin{align*}
2e^t-4 & = 7 \\
2e^t & = 11 \\
e^t & = \frac{11}{2} \\
\end{align*}

So we want to determine the input that gives an output of [latex]\frac{11}{2}[/latex]. We draw the graph of [latex]f(t) = e^t[/latex] and the horizontal line [latex]y = \frac{11}{2}[/latex] in desmos.

Based off the graph, the input is [latex]1.705[/latex]. So, [latex]t \approx 1.705[/latex].

d. We will first do some algebra to get the exponential term by itself on one side so we can use the graph to find a solution:

\begin{align*}
3e^{0.25t}+2 &=6 \\
3e^{0.25t} & = 4 \\
e^{0.25t} & = \frac{4}{3}
\end{align*}

So we want to determine the input that gives an output of [latex]\frac{4}{3}[/latex]. We draw the graph of [latex]f(t) = e^t[/latex] and the horizontal line [latex]y = \frac{4}{3}[/latex] in desmos.

Based off the graph of [latex]e^t[/latex], the input is [latex]0.288[/latex]. In the case of our function the input is [latex]0.25t[/latex]. So, [latex]0.25t = 0.288[/latex]. Dividing both sides by [latex]0.25[/latex] gives [latex]t \approx 1.152.[/latex]

e. We will first do some algebra again to isolate the exponential term:

\begin{align*}
4-2e^{-0.7t} & =3 \\
-2e^{-0.7t} & = -1 \\
e^{-0.7t} & = \frac{1}{2} \\
\end{align*}

So we want to determine the input that gives an output of [latex]\frac{1}{2}[/latex]. We draw the graph of [latex]f(t)=e^t[/latex] and the horizontal line [latex]y = \frac{1}{2}[/latex]in desmos.

Based off the graph of [latex]e^t[/latex], the input is [latex]-0.693[/latex]. In the case of our function the input is [latex]-0.7t[/latex]. So, [latex]-0.7t = -0.693[/latex]. Dividing both sides by [latex]-0.7[/latex] to solve for t we get [latex]t \approx 0.99[/latex].

f. We will again do some algebra to isolate the exponential term:

[latex]2e^{1.2t} = 1.5e^{1.6t}[/latex]

We will first divide both sides by [latex]e^{1.2t}[/latex].

[latex]2 = \frac{1.5e^{1.6t}}{e^{1.2t}}[/latex].

Using exponential rules, when we divide by the same base we subtract the exponents:

[latex]2 = 1.5e^{0.4t}[/latex].

Finally, we divide both sides by 1.5 to get [latex]e^{0.4t} = \frac{2}{1.5} = \frac{2}{3/2} = 2 \cdot \frac{2}{3} = \frac{4}{3}[/latex].

So we want to determine the input that gives an output of [latex]\frac{4}{3}[/latex]. We draw the graph of [latex]f(t) = e^t[/latex] and the horizontal line [latex]y = \frac{4}{3}[/latex] in desmos.

Based off the graph of [latex]e^t[/latex], the input is [latex]0.288[/latex]. In the case of our function our input is [latex]0.4t[/latex]. So, [latex]0.4t = 0.288[/latex]. Dividing both sides by [latex]0.4[/latex] to solve for [latex]t[/latex] we get [latex]t \approx 0.72[/latex].