We are tasked with solving the equation:

\[
0.3 = \frac{5}{3} \left( (x - 1)^{\frac{2}{3}} \right)
\]

### Step-by-step solution:

1. **Isolate the term with \( x \)**:
   Multiply both sides of the equation by \( \frac{3}{5} \) to cancel out the fraction on the right side:

   \[
   \left( \frac{3}{5} \right) \cdot 0.3 = \left( \frac{3}{5} \right) \cdot \frac{5}{3} \left( (x - 1)^{\frac{2}{3}} \right)
   \]
   \[
   0.18 = (x - 1)^{\frac{2}{3}}
   \]

2. **Raise both sides to the power of \( \frac{3}{2} \)**:
   To eliminate the fractional exponent on the right-hand side, raise both sides of the equation to the power of \( \frac{3}{2} \):

   \[
   \left( 0.18 \right)^{\frac{3}{2}} = (x - 1)
   \]

3. **Calculate \( \left( 0.18 \right)^{\frac{3}{2}} \)**:
   We need to calculate \( 0.18^{\frac{3}{2}} \).

   \[
   0.18^{\frac{3}{2}} = (0.18^{1/2})^3 = (\sqrt{0.18})^3
   \]
   \[
   \sqrt{0.18} \approx 0.4243
   \]
   \[
   (0.4243)^3 \approx 0.0763
   \]

   So,

   \[
   x - 1 \approx 0.0763
   \]

4. **Solve for \( x \)**:
   Finally, add 1 to both sides to find \( x \):

   \[
   x = 1 + 0.0763
   \]
   \[
   x \approx 1.0763
   \]

### Final answer:

\[
x \approx 1.0763
\]

Would you like further details or have any questions?

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Here are 5 related questions:

1. How would the solution change if the exponent were \( \frac{3}{2} \) instead of \( \frac{2}{3} \)?
2. What happens if we change the coefficient \( \frac{5}{3} \) to a different value?
3. Can we generalize this method for any fractional exponent in similar equations?
4. How would the solution differ if \( x-1 \) were replaced with \( x+1 \)?
5. What are the steps for solving equations with fractional powers without a calculator?

**Tip:** When solving equations with fractional exponents, raising both sides to the reciprocal of the fraction is a useful technique to eliminate the exponent.

Learn how to solve the equation 0.3 = (5/3)*((x-1)^(2/3)) with a detailed, step-by-step breakdown. This algebra problem covers fractional exponents and exponentiation rules, and is suitable for students in Grades 10-12.

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