Let's solve the equation $9 = 80(1.9)^x$ using logarithms.

Steps:

1. Start by isolating the exponential term: $\frac{9}{80} = (1.9)^x$ Simplifying the left side: $0.1125 = (1.9)^x$

2. Now, apply the natural logarithm (or any logarithm) to both sides: $\ln(0.1125) = \ln((1.9)^x)$

3. Using the logarithmic property $\ln(a^b) = b\ln(a)$, we can simplify the right side: $\ln(0.1125) = x \ln(1.9)$

4. Solve for $x$: $x = \frac{\ln(0.1125)}{\ln(1.9)}$

5. Calculate the values: $\ln(0.1125) \approx -2.1848, \quad \ln(1.9) \approx 0.6419$ So, $x = \frac{-2.1848}{0.6419} \approx -3.4$

Thus, $x \approx -3.4$.

Would you like more details or clarifications on any step?

Here are 5 related questions for further practice:

1. Solve $12 = 100(2.3)^x$ using logarithms.
2. What is the relationship between logarithms and exponents?
3. How do you simplify logarithmic expressions using properties of logarithms?
4. Solve $5^x = 20$ using natural logarithms.
5. How do you apply logarithms to solve exponential growth problems?

Tip: When solving exponential equations, logarithms are extremely useful for "bringing down" exponents and simplifying the equation.