AHL 1.10—Expressions with non-integer exponents

The exponent is $\frac{2}{3}$. The denominator 3 tells us to take the cube root, and the numerator 2 tells us to square the result.

$\sqrt[3]{64}=4\text{since }{4}^3=64$

$4^2=16$

$64^{2/3}=16$

We need $64^{2/3}$. The denominator 3 is the root index, numerator 2 is the power. We use the strategy: root first, then power.

8 would be the result of $64^{1/2}\approx 8$ — but $\sqrt{64}=8$, not the cube root operation combined with squaring. This confuses $\frac{2}{3}$ with $\frac{1}{2}$.

4 is merely $\sqrt[3]{64}$, which is only the first step. The student forgot to apply the numerator power of 2.

32 = $2^5$, which has no direct connection to $64^{2/3}$. This might come from incorrectly multiplying $64\times \frac{1}{2}$.

$\sqrt[3]{64}=4$, then $4^2=16$. The answer is 16.

The rule is: $\sqrt[n]{a^m}=a^{m/n}$. The root index becomes the denominator, and the power becomes the numerator.

Here $n=5$ (the root index) and $m=3$ (the power on $x$).

$\sqrt[5]{x^3}=x^{3/5}$

The correct rational exponent form is $x^{3/5}$.

We need to convert $\sqrt[5]{x^3}$ to exponent form using $a^{m/n}=\sqrt[n]{a^m}$.

$x^{5/3}$ inverts the fraction — it has the root index as the numerator and the power as the denominator. This is the most common reversal error.

$x^{15}$ appears to come from multiplying $5\times 3$. Exponents are not multiplied when converting from radical to rational exponent form.

$x^2$ might come from subtracting $5-3=2$, which has no basis in the conversion rule.

The denominator of the rational exponent is the root index (5) and the numerator is the power (3): $\sqrt[5]{x^3}=x^{3/5}$.