To find the vertical asymptote of the curve:

$y = \frac{65e^{88x} + 110e^{-7x}}{19e^{88x} - 15e^{-7x}}$

We need to determine when the denominator becomes zero because a vertical asymptote occurs when the denominator is zero and the numerator is not zero at the same point.

The denominator of the given function is:

$19e^{88x} - 15e^{-7x} = 0$

Let's solve for $x$:

1. First, rewrite the equation:

$19e^{88x} = 15e^{-7x}$

2. Multiply both sides of the equation by $e^{7x}$ to eliminate the negative exponent:

$19e^{88x} \cdot e^{7x} = 15$

This simplifies to:

$19e^{95x} = 15$

3. Divide both sides by 19:

$e^{95x} = \frac{15}{19}$

4. Take the natural logarithm (ln) of both sides:

$95x = \ln\left(\frac{15}{19}\right)$

5. Solve for $x$:

$x = \frac{1}{95} \ln\left(\frac{15}{19}\right)$

Thus, the vertical asymptote is:

$x = \frac{1}{95} \ln\left(\frac{15}{19}\right)$

Let me know if you'd like further details on these steps or if you have any questions.

Here are 5 related questions you might want to explore:

1. What is the general process for finding vertical asymptotes of rational functions?
2. How does multiplying by $e^{7x}$ help eliminate the negative exponent?
3. What are other types of asymptotes besides vertical, and how do we find them?
4. How do logarithmic properties help solve exponential equations?
5. Can a rational function have multiple vertical asymptotes?

Tip: When solving for vertical asymptotes, always ensure that the numerator is not zero at the same point where the denominator is zero. This ensures a true asymptote rather than a hole in the graph.