Question

$$\frac{A=2^{-4}\times2^{-7}}{2^{-5}\times4^{-5}}$$

Answer

$$A=1/(16384*2^(1/72^5))$$

Solution


Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[A=\frac{1}{{2}^{4}}\times {2}^{-{72}^{-5}}\times {4}^{-5}\]
Simplify  \({2}^{4}\)  to  \(16\).
\[A=\frac{1}{16}\times {2}^{-{72}^{-5}}\times {4}^{-5}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[A=\frac{1}{16}\times \frac{1}{{2}^{{72}^{-5}}}\times {4}^{-5}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[A=\frac{1}{16}\times \frac{1}{{\sqrt[72}^{5]{2}}}\times {4}^{-5}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[A=\frac{1}{16}\times \frac{1}{{\sqrt[72}^{5]{2}}}\times \frac{1}{{4}^{5}}\]
Simplify  \({4}^{5}\)  to  \(1024\).
\[A=\frac{1}{16}\times \frac{1}{{\sqrt[72}^{5]{2}}}\times \frac{1}{1024}\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[A=\frac{1\times 1\times 1}{16{\sqrt[72}^{5]{2}}\times 1024}\]
Simplify  \(1\times 1\)  to  \(1\).
\[A=\frac{1\times 1}{16{\sqrt[72}^{5]{2}}\times 1024}\]
Simplify  \(1\times 1\)  to  \(1\).
\[A=\frac{1}{16{\sqrt[72}^{5]{2}}\times 1024}\]
Simplify  \(16{\sqrt[72}^{5]{2}}\times 1024\)  to  \(16384{\sqrt[72}^{5]{2}}\).
\[A=\frac{1}{16384{\sqrt[72}^{5]{2}}}\]