Solve for x:
$40+30 e^{-.2 x}=50$
The equation given is \(40 + 30e^{-0.2x} = 50\). We are looking for the value of \(x\) that satisfies this equation. The structure of the equation involves a constant term, an exponential term with a negative exponent, and a result. We aim to isolate the exponential part and then solve for \(x\) using logarithmic properties.
To solve for \(x\), we first isolate the term containing the exponential function. Subtracting 40 from both sides of the equation gives:
\[30e^{-0.2x} = 50 - 40\]
\[30e^{-0.2x} = 10\]
Next, divide both sides by 30 to isolate the exponential expression:
\[e^{-0.2x} = \frac{10}{30} = \frac{1}{3}\]
Now, to solve for \(x\) from \(e^{-0.2x} = \frac{1}{3}\), we take the natural logarithm (\(\ln\)) of both sides:
\[\ln(e^{-0.2x}) = \ln\left(\frac{1}{3}\right)\]
Using the property \(\ln(e^u) = u\), the left side simplifies to \(-0.2x\). Using the property \(\ln(\frac{a}{b}) = \ln(a) - \ln(b)\), the right side becomes \(\ln(1) - \ln(3)\). Since \(\ln(1) = 0\), we have:
\[-0.2x = -\ln(3)\]
Dividing both sides by -0.2 gives:
\[x = \frac{-\ln(3)}{-0.2} = \frac{\ln(3)}{0.2}\]
Based on our derivation, the value of \(x\) is \(\frac{\ln(3)}{0.2}\). Comparing this result to the given options, it matches the final choice. Thus, the correct expression is \(\boxed{\frac{\ln(3)}{0.2}}\).
Exponential functions are fundamental in modeling real-world processes where growth or decay occurs at a rate proportional to the current value. A classic example is radioactive decay, where a material decreases over time following a formula like \(N(t) = N_0 e^{-\lambda t}\). By solving equations similar to the one above, scientists can determine the "half-life" of a substance, which is the time it takes for half of the radioactive atoms to decay. Understanding how to isolate variables in these exponential equations is essential for accurate predictions in physics, biology, and finance.