Solve for p p - 4 = -9+ p
2 answers:
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Megan:
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>![\sqrt[32]{x* x^{2} * x^{5} }](https://tex.z-dn.net/?f=%20%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%20)
<span>
Since </span>
Then![\sqrt[32]{x* x^{2} * x^{5} }= \sqrt[32]{ x^{1+2+5} } =\sqrt[32]{ x^{8} }](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7Bx%2A%20x%5E%7B2%7D%20%2A%20x%5E%7B5%7D%20%7D%3D%20%5Csqrt%5B32%5D%7B%20x%5E%7B1%2B2%2B5%7D%20%7D%20%3D%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D)
Since![\sqrt[n]{x^{m}} = x^{m/n} }](https://tex.z-dn.net/?f=%20%5Csqrt%5Bn%5D%7Bx%5E%7Bm%7D%7D%20%3D%20x%5E%7Bm%2Fn%7D%20%7D%20)
Then![\sqrt[32]{ x^{8} }= x^{8/32} = x^{1/4}](https://tex.z-dn.net/?f=%5Csqrt%5B32%5D%7B%20x%5E%7B8%7D%20%7D%3D%20x%5E%7B8%2F32%7D%20%3D%20x%5E%7B1%2F4%7D%20)
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.
x to the one third power =
<span>x to the one twelfth power = </span>
<span>The quantity of x to the one third power, over x to the one twelfth power is:
</span>
<span>
Since </span>
then
Now, just subtract exponents:
1/3 - 1/12 = 4/12 - 1/12 = 3/12 = 1/4
Julie:
x times x to the second times x to the fifth = x * x² * x⁵
<span>The thirty second root of the quantity of x times x to the second times x to the fifth is
</span>
<span>
Since </span>
Then
Since
Then
Since both Megan and Julie got the same result, it can be concluded that their expressions are equivalent.