Question

simplify. assume all variables are positive.
$(36u^{7})^{-\frac{1}{2}}$
write your answer in the form $a$ or $\frac{a}{b}$, where $a$ and $b$ are constants or variable expressions that have no variables in common. all exponents in your answer should be positive.

Explanation:

Step1: Apply the power of a product rule

The power of a product rule states that \((ab)^n = a^n b^n\). So, we can apply this to \((36u^7)^{-\frac{1}{2}}\) as follows:
\((36u^7)^{-\frac{1}{2}} = 36^{-\frac{1}{2}} \cdot (u^7)^{-\frac{1}{2}}\)

Step2: Simplify the constant term

Recall that \(a^{-\frac{1}{2}}=\frac{1}{a^{\frac{1}{2}}}=\frac{1}{\sqrt{a}}\). For \(36^{-\frac{1}{2}}\), we have:
\(36^{-\frac{1}{2}}=\frac{1}{36^{\frac{1}{2}}}=\frac{1}{\sqrt{36}}=\frac{1}{6}\)

Step3: Simplify the variable term

Using the power of a power rule \((a^m)^n = a^{mn}\), for \((u^7)^{-\frac{1}{2}}\) we get:
\((u^7)^{-\frac{1}{2}} = u^{7\times(-\frac{1}{2})}=u^{-\frac{7}{2}}\)
And since \(a^{-n}=\frac{1}{a^n}\), we can rewrite \(u^{-\frac{7}{2}}\) as \(\frac{1}{u^{\frac{7}{2}}}\)

Step4: Combine the constant and variable terms

Now, multiply the results from Step 2 and Step 3:
\(36^{-\frac{1}{2}} \cdot (u^7)^{-\frac{1}{2}}=\frac{1}{6}\cdot\frac{1}{u^{\frac{7}{2}}}=\frac{1}{6u^{\frac{7}{2}}}\)

Answer:

\(\frac{1}{6u^{\frac{7}{2}}}\)