We are given the equation:

$103.5 = e^{\left(-7966.2\left(\frac{1}{x}\right) + 21.1\right)}$

To solve for $x$, follow these steps:

Step 1: Take the natural logarithm (ln) of both sides

Since the natural logarithm and the exponential function are inverse operations, applying $\ln$ to both sides will simplify the equation:

$\ln(103.5) = \ln\left(e^{\left(-7966.2\left(\frac{1}{x}\right) + 21.1\right)}\right)$

$\ln(103.5) = -7966.2\left(\frac{1}{x}\right) + 21.1$

Step 2: Simplify $\ln(103.5)$

We can now compute the value of $\ln(103.5)$:

$\ln(103.5) \approx 4.638$

So, the equation becomes:

$4.638 = -7966.2\left(\frac{1}{x}\right) + 21.1$

Step 3: Isolate the term with $\frac{1}{x}$

Subtract 21.1 from both sides:

$4.638 - 21.1 = -7966.2\left(\frac{1}{x}\right)$

$-16.462 = -7966.2\left(\frac{1}{x}\right)$

Step 4: Solve for $\frac{1}{x}$

Divide both sides by -7966.2:

$\frac{1}{x} = \frac{-16.462}{-7966.2} \approx 0.002066$

Step 5: Solve for $x$

Take the reciprocal of both sides:

$x \approx \frac{1}{0.002066} \approx 484.0$

Thus, the solution is:

$x \approx 484.0$

Let me know if you would like any further clarifications!

Related Questions:

1. How do we verify the solution by plugging $x$ back into the equation?
2. What happens if the base of the exponent is not $e$?
3. What if there was a different constant in front of the exponential expression?
4. Can we solve similar equations if the exponent includes logarithms instead of reciprocals?
5. How does the graph of this equation behave as $x$ approaches large values?

Tip:

Always isolate the exponential term before applying the logarithm for simpler calculations.