Maths / Exponents and Power / Laws of exponent when exponent is - n

QUESTION

Simplify:

a.  $\dpi{150} \large \frac{4}{(216)^\frac{-2}{3}}+\frac{1}{(256)^\frac{-3}{4}}+\frac{2}{(243)^\frac{-1}{5}}$

b. $\dpi{150} \large \frac{16\times2^{n+1}-4\times2^n}{16 \times2^{n+2}-2\times2^{n+2}}$

EXPLANATION
Explain TypeExplanation Content
Text
• $\dpi{150} \large \frac{4}{(216)^\frac{-2}{3}}+\frac{1}{(256)^\frac{-3}{4}}+\frac{2}{(243)^\frac{-1}{5}}$

$\dpi{150} \large =\frac{4}{(6^3)^\frac{-2}{3}}+\frac{1}{(4^4)^\frac{-3}{4}}+\frac{2}{(3^5)^\frac{-1}{5}}$

$\dpi{150} \large =\frac{4}{(6^{-2})}+\frac{1}{(4^{-3})}+\frac{2}{(3^{-1})}$

$\dpi{150} \large =4\times 36 +64+6$

$\dpi{150} \large =214$

• $\dpi{150} \large \frac{16\times2^{n+1}-4\times2^n}{16 \times2^{n+2}-2\times2^{n+2}}$

$\dpi{120} \fn_jvn \large =\large\frac{4\times 2^n(4\times 2-1)}{2\times 2^{n+2}(8-1)}$

$\dpi{120} \fn_jvn \large =\large\frac{2\times (8-1)}{2^2(8-1)}$

$\dpi{120} \fn_jvn \large =\large\frac{1}{2}$

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