Year 11 (2026) Maths Advanced (2026) Logs and Exponentials

Index Laws (Integer Indices)

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Questions

Question 1

124513

Simplify \(\dfrac{4^n \times 8^{n+1}}{16^{n-1}}\)

Answer

\(2^{n+7}\)

Worked Solution

\begin{aligned}
\frac{4^n \times 8^{n+1}}{16^{n-1}}&=\frac{\left(2^2\right)^n \times\left(2^3\right)^{n+1}}{\left(2^4\right)^{n-1}} \\
& =\frac{2^{2 n} \times 2^{3 n+3}}{2^{4 n-4}} \\
& =2^{5 n+3-(4 n-4)} \\
& =2^{n+7}
\end{aligned}

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