Question

all exponential functions can be written in many forms. write the function $f(t) = 78000e^{0.15t}$ in the form $f(t) = ab^{5t}$. round all coefficients to four decimal places.

Explanation:

Step1: Set the two forms equal

We have \(78000e^{0.15t}=ab^{5t}\).

Step2: Express \(e^{0.15t}\) as a power of \(b^{5t}\)

Let's rewrite \(e^{0.15t}\) in terms of \(b^{5t}\). We know that \(e^{0.15t}=(e^{0.15/5})^{5t}\) (using the property \((x^{m})^{n}=x^{mn}\), here \(x = e\), \(m=0.15/5\), \(n = 5t\)).

Step3: Find the value of \(b\)

So we can set \(b = e^{0.15/5}\). Calculate \(0.15/5=0.03\), then \(b = e^{0.03}\approx1.030454\). Rounding to four decimal places, \(b\approx1.0305\). And \(a = 78000\) since when we compare \(78000e^{0.15t}=ab^{5t}\) and \(78000(e^{0.03})^{5t}=ab^{5t}\), we see that \(a = 78000\) and \(b = e^{0.03}\).

Step4: Verify the form

Now we can write \(f(t)=78000\times(1.0305)^{5t}\). Let's check: \((1.0305)^{5t}=e^{\ln(1.0305)\times5t}\), and \(\ln(1.0305)\approx0.03\), so \(5\times0.03 = 0.15\), which matches the exponent in the original function.

Answer:

\(f(t)=78000\times1.0305^{5t}\) (where \(a = 78000\) and \(b\approx1.0305\))