Question

I need help with a few exponent problems. I have listed three, but even help with just one would be helpful. I already did the first one and just need someone to check it. I included typed and written versions of the questions.
2. 4x^{2/4}/ 8x^{1/3}
3. a^{5/6} b^{2/6}
4. (m*m^{-2}n^{1/3})^{2}

Answer 1

(In your typed questions, you forgot to make the 2/4 negative, I had to go back and redo once I realized lol.)

To solve the second problem, you must use the quotient of a power property. The quotient of a power property states that exponents of a division problem are subtracted. That means that -2/4 and 1/3 will be subtracted. To do this, we must find common denominators. The best one to use would be 12, so that is what we will do. Remember to multiply the numerator by whatever number you did to the denominator to get 12. This causes -2/4 to become -6/12, and 1/3 to become 4/12. Then, subtract them to get -10/12. This answer can be simplified down to -5/6.

For the third problem, you solve using the power product property. This property states that when multiplying, add the exponents. Since they already have common denominators, you can skip that step. When you add them it comes out to 8/6. You simplify that, and if fully simplified, it comes out to 1 1/3.

If you want me to do the others I can, just comment below, and I will edit my answer to include them.

Answer 2

(2)

$\frac{4 x^{-2/4}}{8 x^{1/3}}$

First simplify -2/4 dividing each term by 2

$\frac{-2/2}{4/2} = -1/2$

Then, combine x term from numerator with x term from denominator, and non-x term from numerator with non-x term from denominator

$\frac{4 x^{-2/4}}{8 x^{1/3}}$

$\frac{4 x^{-1/2}}{8 x^{1/3}}$

$\frac{4}{8} \frac{x^{-1/2}}{x^{1/3}}$

$\frac{4}{8} x^{(-1/2 - 1/3)}$

$\frac{1}{2} x^{-5/6}$

(3)

$a^{6/6} b^{2/6}$

6/6 = 1 and 2/6 can be simplified dividing numertor and denominator by 2, which gives 2/6 = 1/3. Therefore

$a^{6/6} b^{2/6} = a^1 b^{1/3} = a \sqrt[3]{b}$

(4)

${(m m^{-2} n^{1/3})}^2$

Negative exponents can be expressed as a fraction in this way

$m^{-2} = \frac{1}{m^2}$

Replacing it in the equation gives

${(\frac{m}{m^2} n^{1/3})}^2$

After m simplification

${(\frac{n^{1/3}}{m})}^2$

$\frac{(n^{1/3})^2}{m^2}$

$\frac{n^{(1/3\times2)}}{m^2}$

$\frac{n^{2/3}}{m^2}$