Question

$$\frac{27^{1}\frac{1}{3}\times3b^{-}}{3b^{-}5}$$

Answer

$$(16.642808065898*10^161)/b^8$$

Solution


Simplify  \({27}^{113}\)  to  \(5.547603\times {10}^{161}\).
\[5.547603\times {10}^{161}\times 3{b}^{-3}{b}^{-5}\]
Take out the constants.
\[(5.547603\times 3){b}^{-3}{b}^{-5}\times {10}^{161}\]
Simplify  \(5.547603\times 3\)  to  \(16.642808\).
\[16.642808{b}^{-3}{b}^{-5}\times {10}^{161}\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[16.642808{b}^{-3-5}\times {10}^{161}\]
Simplify  \(-3-5\)  to  \(-8\).
\[16.642808{b}^{-8}\times {10}^{161}\]
Use Negative Power Rule: \({x}^{-a}=\frac{1}{{x}^{a}}\).
\[16.642808\times \frac{1}{{b}^{8}}\times {10}^{161}\]
Use this rule: \(\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}\).
\[\frac{16.642808\times 1\times {10}^{161}}{{b}^{8}}\]
Simplify  \(16.642808\times 1\)  to  \(16.642808\).
\[\frac{16.642808\times {10}^{161}}{{b}^{8}}\]