Let's go through each equation step by step, showing all the work, and then address which equation has multiple solutions.

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1) Solve the equation:

$(4x + 1)^{\frac{3}{2}} = 27$

Step 1: Isolate the base term with the exponent.

$(4x + 1)^{\frac{3}{2}} = 27$ To eliminate the exponent $\frac{3}{2}$, raise both sides of the equation to the power of $\frac{2}{3}$: $\left( (4x + 1)^{\frac{3}{2}} \right)^{\frac{2}{3}} = 27^{\frac{2}{3}}$ This simplifies to: $4x + 1 = 27^{\frac{2}{3}}$ Now, calculate $27^{\frac{2}{3}}$. Since $27 = 3^3$, the fractional exponent $\frac{2}{3}$ gives: $27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^2 = 9$ Thus, we have: $4x + 1 = 9$

Step 2: Solve for $x$.

Subtract 1 from both sides: $4x = 8$ Now, divide by 4: $x = 2$

So, the solution is: $x = 2$

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2) Solve the equation:

$4x^{\frac{2}{3}} = 9$

Step 1: Isolate the term with the exponent.

Divide both sides by 4: $x^{\frac{2}{3}} = \frac{9}{4}$

Step 2: Eliminate the exponent $\frac{2}{3}$.

Raise both sides to the power of $\frac{3}{2}$ to cancel out the fractional exponent: $\left( x^{\frac{2}{3}} \right)^{\frac{3}{2}} = \left( \frac{9}{4} \right)^{\frac{3}{2}}$ This simplifies to: $x = \left( \frac{9}{4} \right)^{\frac{3}{2}}$

Step 3: Simplify the right-hand side.

To compute $\left( \frac{9}{4} \right)^{\frac{3}{2}}$, break it down: $\left( \frac{9}{4} \right)^{\frac{3}{2}} = \left( \frac{9}{4} \right)^1.5 = \left( \frac{9}{4} \right)^1 \times \left( \frac{9}{4} \right)^{\frac{1}{2}}$ $= \frac{9}{4} \times \frac{3}{2} = \frac{27}{8}$ Thus, we have: $x = \frac{27}{8}$

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3) Which equation has multiple solutions?

To determine which equation has multiple solutions, we analyze the nature of the powers and exponents involved.

• In Equation 1, $(4x + 1)^{\frac{3}{2}} = 27$, the base expression $(4x + 1)$ is raised to a fractional exponent greater than 1. However, the cube root operation is well-behaved, and solving for $x$ leads to a single value: $x = 2$. No multiple solutions exist here because the expression is a one-to-one function for real values of $x$.

• In Equation 2, $4x^{\frac{2}{3}} = 9$, the variable $x$ is raised to the fractional power $\frac{2}{3}$. Since raising a number to an even root (such as the square root) introduces the possibility of both positive and negative values, this equation has multiple solutions. The exponent $\frac{2}{3}$ involves squaring, which means both positive and negative values of $x$ can satisfy the equation after raising to the power of $\frac{3}{2}$.

Hence, Equation 2 has multiple solutions because of the even root involved.

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Summary:

• The solution to the first equation is $x = 2$.
• The solution to the second equation is $x = \frac{27}{8}$, but it has multiple solutions because it involves an even root.

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Let me know if you'd like any details or clarifications!
Here are some related questions to expand your understanding:

1. What would happen if the exponent in the first equation was $\frac{1}{2}$ instead of $\frac{3}{2}$?
2. How do fractional exponents like $\frac{2}{3}$ affect the number of solutions in equations?
3. How do we handle equations with cube roots differently from square roots?
4. What happens when you square both sides of an equation with negative numbers?
5. How would you graph the solutions to these equations?

Tip: When solving equations with fractional exponents, pay special attention to even roots as they can introduce multiple solutions.