- Divide both sides by 8 to obtain \(e^{2x+1} = 0.5\).
- Apply the natural logarithm to both sides to get \(2x + 1 = \ln(0.5)\).
- Subtract 1 and divide by 2 to isolate \(x\).
- The solution is \(x = \frac{\ln(0.5)-1}{2}\), which corresponds to the fourth option: \(\boxed{x = \frac{\ln(0.5)-1}{2}}\).
Explanation
- Analysis of the Equation
The equation provided is \(8(e)^{2x+1} = 4\), where \(e\) is the base of the natural logarithm. Our objective is to isolate the variable \(x\). To solve this, we will apply the properties of exponents and logarithms, specifically isolating the exponential term and then applying the natural logarithm function to both sides.
- Solving for x
First, we isolate the exponential expression \(e^{2x+1}\) by dividing both sides of the equation by 8:
\[e^{2x+1} = \frac{4}{8}\]
\[e^{2x+1} = 0.5\]
Next, to solve for \(x\) in the exponent, we take the natural logarithm (\(\ln\)) of both sides:
\[\ln(e^{2x+1}) = \ln(0.5)\]
Using the property \(\ln(e^y) = y\), we simplify the left side:
\[2x + 1 = \ln(0.5)\]
Now, we subtract 1 from both sides:
\[2x = \ln(0.5) - 1\]
Finally, divide by 2 to isolate \(x\):
\[x = \frac{\ln(0.5) - 1}{2}\]
- Conclusion
Comparing our derived solution \(x = \frac{\ln(0.5) - 1}{2}\) to the provided options, we can see that it matches the fourth choice:
- \(x = \frac{\ln(0.5)+1}{2}\)
- \(x = \frac{\ln(0.5)}{2}+1\)
- \(x = \frac{\ln(0.5)}{2}-1\)
- \(x = \frac{\ln(0.5)-1}{2}\)
Therefore, the correct solution is the fourth option, \(x = \frac{\ln(0.5)-1}{2}\).
Examples
Understanding how to solve exponential equations with the base \(e\) is essential in various fields such as finance, physics, and biology. For instance, if you are modeling the cooling of an object or the decay of a substance, you often encounter expressions like \(N(t) = N_0 e^{kt}\). Solving for the time \(t\) requires taking the natural logarithm, just as we did here, allowing you to determine exactly when a certain amount of the substance will remain. This skill is a fundamental tool for working with exponential growth and decay models in real-world scenarios.