SL 1.7—Laws of exponents and logs

We have $27^{\frac{4}{3}}$, where the denominator $n=3$ gives the root and the numerator $m=4$ gives the power.

It is easier to take the root first: $\sqrt[3]{27}=3$, since $3^3=27$.

$27^{\frac{4}{3}}={\left(\sqrt[3]{27}\right)}^4=3^4=81$

The answer is 81.

We need to evaluate $27^{\frac{4}{3}}$, which means the cube root of 27, raised to the 4th power.

12 would result from $27\times \frac{4}{9}$ or some incorrect arithmetic — not from correctly applying the rational exponent definition.

36 might come from confusing $27^{\frac{4}{3}}$ with $9^2=81$... wait, or perhaps multiplying $27\times \frac{4}{3}=36$, which incorrectly treats the exponent as a multiplier.

108 could result from computing $27\times 4=108$, incorrectly treating the exponent as multiplication rather than applying root and power.

Correctly: $\sqrt[3]{27}=3$, then $3^4=81$. The answer is 81.

Using $a^x\cdot a^y=a^{x+y}$: $x^{\frac{5}{4}}\cdot x^{\frac{3}{4}}=x^{\frac{5}{4}+\frac{3}{4}}=x^{\frac{8}{4}}=x^2$

$\frac{x^2}{x^{\frac{1}{2}}}=x^{2-\frac{1}{2}}=x^{\frac{4}{2}-\frac{1}{2}}=x^{\frac{3}{2}}$

The simplified expression is $x^{\frac{3}{2}}$.

We must combine exponents using the product and quotient rules: add exponents when multiplying, subtract when dividing.

$x$ corresponds to exponent 1. The numerator exponent sum is $\frac{5}{4}+\frac{3}{4}=2$, and $2-\frac{1}{2}=\frac{3}{2}\ne 1$, so this is incorrect.

$x^2$ ignores the division by $x^{\frac{1}{2}}$. The numerator alone gives $x^2$, but dividing reduces the exponent.

$x^{\frac{7}{4}}$ might come from only adding $\frac{5}{4}+\frac{3}{4}-\frac{1}{2}=\frac{8}{4}-\frac{2}{4}=\frac{6}{4}$... actually $\frac{7}{4}$ could come from adding all three exponents incorrectly.

Numerator: $\frac{5}{4}+\frac{3}{4}=2$; then $2-\frac{1}{2}=\frac{3}{2}$. The answer is $x^{\frac{3}{2}}$.