$$ 3^{x\text{ }} $$$$ \cdot4^{x+1} $$$$ =5^{x+2} $$

To solve the equation:

\[
3^x \cdot 4^{x+1} = 5^{x+2}
\]

we can follow these steps:

1. **Rewrite the equation** to isolate the exponential terms:

\[
3^x \cdot 4^x \cdot 4 = 5^x \cdot 25
\]

This simplifies to:

\[
(3^x \cdot 4^x) \cdot 4 = 5^x \cdot 25
\]

2. **Combine the exponents on the left-hand side**:

\[
(3 \cdot 4)^x \cdot 4 = 5^x \cdot 25
\]

\[
12^x \cdot 4 = 5^x \cdot 25
\]

3. **Divide both sides by** $4$:

\[
12^x = \frac{25 \cdot 5^x}{4}
\]

4. **Express the equation in terms of logarithms**:

Take the logarithm on both sides:

\[
\log(12^x) = \log\left(\frac{25 \cdot 5^x}{4}\right)
\]

Using the logarithm properties:

\[
x \log(12) = \log(25) + x \log(5) - \log(4)
\]

5. **Isolate** $x$:

\[
x \log(12) - x \log(5) = \log(25) - \log(4)
\]

\[
x(\log(12) - \log(5)) = \log\left(\frac{25}{4}\right)
\]

\[
x = \frac{\log\left(\frac{25}{4}\right)}{\log\left(\frac{12}{5}\right)}
\]

6. **Calculate** the values:

\[
x \approx \frac{\log(6.25)}{\log(2.4)}
\]

Using the logarithm values:

\[
x \approx \frac{0.79588}{0.38021} \approx 2.093
\]

So, the solution is:

\[
x \approx 2.093
\]

Would you like more details on any step, or do you have any questions?

---

Here are some related questions:

1. How do logarithmic properties help in solving exponential equations?
2. What are some alternative methods to solve equations involving exponents?
3. How does the change of base formula work in logarithms?
4. What is the significance of the logarithm function in solving exponential equations?
5. Can this type of problem be solved graphically? If so, how?

**Tip:** When solving exponential equations, converting to logarithmic form is a powerful tool, especially when the bases are different.

Learn how to solve the exponential equation 3^x * 4^(x+1) = 5^(x+2) with a step-by-step solution involving logarithms and exponential properties. Suitable for advanced high school students or those studying logarithms and exponential equations.

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