Sure, let's solve the equation step-by-step:
Given equation:
[tex]\[ 240 = 1920 e^{-0.5x} \][/tex]
1. Isolate the exponential term:
Divide both sides of the equation by 1920:
[tex]\[
\frac{240}{1920} = e^{-0.5x}
\][/tex]
Simplify the fraction:
[tex]\[
0.125 = e^{-0.5x}
\][/tex]
2. Take the natural logarithm on both sides:
To get rid of the exponential, take the natural logarithm ([tex]\(\ln\)[/tex]) of both sides:
[tex]\[
\ln(0.125) = \ln(e^{-0.5x})
\][/tex]
Using the property [tex]\(\ln(e^y) = y\)[/tex], we have:
[tex]\[
\ln(0.125) = -0.5x
\][/tex]
3. Solve for [tex]\(x\)[/tex]:
To isolate [tex]\(x\)[/tex], divide both sides by -0.5:
[tex]\[
x = \frac{\ln(0.125)}{-0.5}
\][/tex]
4. Express the exact value:
Recognize that [tex]\(0.125 = \frac{1}{8}\)[/tex], and using the property of logarithms [tex]\(\ln\left(\frac{1}{a}\right) = -\ln(a)\)[/tex]:
[tex]\[
\ln(0.125) = \ln\left(\frac{1}{8}\right) = -\ln(8)
\][/tex]
Substitute this back into the expression:
[tex]\[
x = \frac{-\ln(8)}{-0.5} = \frac{\ln(8)}{0.5} = 2 \ln(8)
\][/tex]
5. Approximate the solution to 4 decimal places:
Using the numerical approximation of the natural logarithm, [tex]\(\ln(8) \approx 2.0794\)[/tex]:
[tex]\[
x \approx 2 \cdot 2.0794 = 4.1588
\][/tex]
Round to 4 decimal places:
[tex]\[
x \approx 4.1589
\][/tex]
So, the exact solution is:
[tex]\[ x = 2 \ln 8 \][/tex]
And the approximate solution to 4 decimal places is:
[tex]\[ x \approx 4.1589 \][/tex]
Thus, the solution set with the exact value in terms of natural logarithms and the approximate solution to 4 decimal places is:
[tex]\[
x = 2 \ln 8 \quad \text{(exact solution)}
\][/tex]
[tex]\[
x \approx 4.1589 \quad \text{(approximate solution to 4 decimal places)}
\][/tex]
