To solve the equation $10e^{2x-3} = 15e^{5x-7}$, follow these steps:

1. Rewrite the equation to isolate the exponential terms: $\frac{10e^{2x-3}}{15e^{5x-7}} = 1$

2. Simplify the fraction: $\frac{10}{15} \cdot \frac{e^{2x-3}}{e^{5x-7}} = 1$ $\frac{2}{3} \cdot e^{(2x-3) - (5x-7)} = 1$

3. Combine the exponents: $e^{(2x-3) - (5x-7)} = e^{2x-3 - 5x + 7} = e^{-3x + 4}$ $\frac{2}{3} \cdot e^{-3x + 4} = 1$

4. Isolate the exponential expression: $e^{-3x + 4} = \frac{3}{2}$

5. Take the natural logarithm (ln) of both sides to solve for $x$: $-3x + 4 = \ln\left(\frac{3}{2}\right)$

6. Solve for $x$: $-3x = \ln\left(\frac{3}{2}\right) - 4$ $x = \frac{4 - \ln\left(\frac{3}{2}\right)}{3}$

Therefore, the solution to the equation $10e^{2x-3} = 15e^{5x-7}$ is: $x = \frac{4 - \ln\left(\frac{3}{2}\right)}{3}$

Would you like any further details or have any questions?

Here are some questions you might want to ask next:

1. How do you solve logarithmic equations?
2. What are the properties of exponents?
3. Can you explain the natural logarithm function?
4. How do you solve quadratic equations?
5. What is the derivative of an exponential function?

Tip: When solving exponential equations, taking the natural logarithm of both sides can often simplify the process and help isolate the variable.