To solve the equation:

\[
50 \cdot 3^{3x} = 9
\]

we need to isolate the exponential expression involving \( x \). Let's follow these steps:

### Step 1: Divide both sides by 50

\[
3^{3x} = \frac{9}{50}
\]

### Step 2: Express 9 as a power of 3

Since \( 9 = 3^2 \), the equation becomes:

\[
3^{3x} = \frac{3^2}{50}
\]

### Step 3: Apply the logarithm

Take the natural logarithm (or common logarithm) on both sides:

\[
\ln(3^{3x}) = \ln\left(\frac{3^2}{50}\right)
\]

By the logarithmic property \( \ln(a^b) = b \ln(a) \), we have:

\[
3x \ln(3) = \ln(3^2) - \ln(50)
\]

### Step 4: Simplify the right-hand side

\[
3x \ln(3) = 2 \ln(3) - \ln(50)
\]

### Step 5: Solve for \( x \)

Divide both sides by \( 3 \ln(3) \):

\[
x = \frac{2 \ln(3) - \ln(50)}{3 \ln(3)}
\]

### Step 6: Numerical approximation

To get a numerical value, use the approximations \( \ln(3) \approx 1.0986 \) and \( \ln(50) \approx 3.912 \):

\[
x \approx \frac{2 \cdot 1.0986 - 3.912}{3 \cdot 1.0986}
\]

Calculating this:

\[
x \approx \frac{2.1972 - 3.912}{3.2958} \approx \frac{-1.7148}{3.2958} \approx -0.5202
\]

### Solution

\[
x \approx -0.5202
\]

Would you like more details or have any other questions?

### Related Questions:

1. How do logarithmic functions work, and why are they useful in solving exponential equations?
2. Can you explain the properties of logarithms in more depth?
3. How can I solve exponential equations without logarithms?
4. What are some real-life applications of exponential and logarithmic functions?
5. Can we use different bases for logarithms to solve this equation?

### Tip:

Remember that logarithms are the inverse of exponentiation and are extremely useful for solving equations where the variable is in the exponent.

Learn how to solve the exponential equation 50⋅3^3x=9 using logarithms. This detailed step-by-step solution covers key concepts such as exponential equations, logarithmic properties, and solving for unknowns in the exponent. Suitable for students in Grades 9-12.

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