Answer :
To solve the equation [tex]\(6 \cdot e^{0.25 t} = 9\)[/tex] for [tex]\(t\)[/tex], follow these steps:
1. Isolate the exponential expression:
[tex]\[ 6 \cdot e^{0.25 t} = 9 \][/tex]
Divide both sides of the equation by 6:
[tex]\[ e^{0.25 t} = \frac{9}{6} \][/tex]
Simplify the fraction:
[tex]\[ e^{0.25 t} = 1.5 \][/tex]
2. Take the natural logarithm of both sides:
By applying the natural logarithm (denoted as [tex]\(\ln\)[/tex]) to both sides, we have:
[tex]\[ \ln(e^{0.25 t}) = \ln(1.5) \][/tex]
3. Utilize the properties of logarithms:
Recall that [tex]\(\ln(e^x) = x\)[/tex]. Therefore:
[tex]\[ 0.25 t = \ln(1.5) \][/tex]
4. Solve for [tex]\(t\)[/tex]:
To isolate [tex]\(t\)[/tex], divide both sides by 0.25:
[tex]\[ t = \frac{\ln(1.5)}{0.25} \][/tex]
Hence, the exact expression for [tex]\(t\)[/tex] is:
[tex]\[ t = \frac{\ln(1.5)}{0.25} \][/tex]
5. Approximate the value of [tex]\(t\)[/tex]:
Using a calculator to evaluate [tex]\(\frac{\ln(1.5)}{0.25}\)[/tex]:
[tex]\[ t \approx 1.622 \][/tex]
So, your final answers are:
[tex]\[ t = \frac{\ln(1.5)}{0.25} \][/tex]
[tex]\[ t \approx 1.622 \][/tex]
1. Isolate the exponential expression:
[tex]\[ 6 \cdot e^{0.25 t} = 9 \][/tex]
Divide both sides of the equation by 6:
[tex]\[ e^{0.25 t} = \frac{9}{6} \][/tex]
Simplify the fraction:
[tex]\[ e^{0.25 t} = 1.5 \][/tex]
2. Take the natural logarithm of both sides:
By applying the natural logarithm (denoted as [tex]\(\ln\)[/tex]) to both sides, we have:
[tex]\[ \ln(e^{0.25 t}) = \ln(1.5) \][/tex]
3. Utilize the properties of logarithms:
Recall that [tex]\(\ln(e^x) = x\)[/tex]. Therefore:
[tex]\[ 0.25 t = \ln(1.5) \][/tex]
4. Solve for [tex]\(t\)[/tex]:
To isolate [tex]\(t\)[/tex], divide both sides by 0.25:
[tex]\[ t = \frac{\ln(1.5)}{0.25} \][/tex]
Hence, the exact expression for [tex]\(t\)[/tex] is:
[tex]\[ t = \frac{\ln(1.5)}{0.25} \][/tex]
5. Approximate the value of [tex]\(t\)[/tex]:
Using a calculator to evaluate [tex]\(\frac{\ln(1.5)}{0.25}\)[/tex]:
[tex]\[ t \approx 1.622 \][/tex]
So, your final answers are:
[tex]\[ t = \frac{\ln(1.5)}{0.25} \][/tex]
[tex]\[ t \approx 1.622 \][/tex]
