https://file.aiquickdraw.com/mathbot/file/1728525368491fq85xmad.jpg

The equation from the image is:

\[
f(x) = 5(2.2)^x \quad \text{and} \quad g(x) = 58(1.5)^x
\]
We are tasked with solving \( f(x) = g(x) \) for \( x \). 

This means we set the two functions equal to each other:

\[
5(2.2)^x = 58(1.5)^x
\]

### Steps to solve:

1. Divide both sides by \( (1.5)^x \) to isolate \( (2.2)^x \) and \( (1.5)^x \):
   \[
   \frac{(2.2)^x}{(1.5)^x} = \frac{58}{5}
   \]
   This simplifies to:
   \[
   \left( \frac{2.2}{1.5} \right)^x = \frac{58}{5}
   \]
   Calculate the right-hand side:
   \[
   \frac{58}{5} = 11.6
   \]
   So we have:
   \[
   \left( \frac{2.2}{1.5} \right)^x = 11.6
   \]
   
2. Now take the natural logarithm of both sides:
   \[
   \ln\left(\left(\frac{2.2}{1.5}\right)^x\right) = \ln(11.6)
   \]
   Using the property \( \ln(a^x) = x \ln(a) \), this becomes:
   \[
   x \ln\left(\frac{2.2}{1.5}\right) = \ln(11.6)
   \]
   
3. Solve for \( x \):
   \[
   x = \frac{\ln(11.6)}{\ln\left(\frac{2.2}{1.5}\right)}
   \]

### Calculating the values:
- \( \frac{2.2}{1.5} \approx 1.4667 \)
- \( \ln(11.6) \approx 2.451 \)
- \( \ln(1.4667) \approx 0.383 \)

Thus:
\[
x = \frac{2.451}{0.383} \approx 6.4
\]

### Final Answer:
The solution is \( x \approx 6.4 \).

Would you like more details on any of these steps, or have any questions?

Here are some related questions to expand on this:
1. How do you interpret logarithms in exponential equations?
2. What other methods can you use to solve exponential equations?
3. How would the equation change if different bases were used in \( f(x) \) or \( g(x) \)?
4. What is the significance of the base in exponential growth functions?
5. Can you apply logarithmic properties to simplify more complex expressions?

**Tip**: Always use natural logarithms when dealing with exponential equations involving \( e^x \), but common logarithms (base 10) can work well for other bases.

Suppose f(x) = 5(2.2)^x and g(x) = 58(1.5)^x. Solve f(x) = g(x) for x.

Learn how to solve the exponential equation 5(2.2)^x = 58(1.5)^x using logarithmic properties. This problem covers concepts like exponents and logarithms, and it is suitable for high school students (Grades 10-12).

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