Simplify the following:$ \frac{6(8)^{n+1}+16(2)^{3 n-2}}{10(2)^{3 n+1}-7(8)^{n}} $

Given:

\( \frac{6(8)^{n+1}+16(2)^{3 n-2}}{10(2)^{3 n+1}-7(8)^{n}} \)

To do:

We have to simplify the given expression.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$  

$\frac{6(8)^{n+1}+16(2)^{3 n-2}}{10(2)^{3 n+1}-7(8)^{n}}=\frac{6(2^{3})^{n+1}+16(2)^{3 n-2}}{10(2)^{3 n+1}-7(2^{3})^{n}}$

$=\frac{6 \times 2^{3 n+3}+16 \times 2^{3 n-2}}{10\times2^{3 n+1}-7 \times 2^{3 n}}$

$=\frac{2^{3 n}(6 \times 2^{3}+16 \times 2^{-2})}{2^{3 n}(10 \times 2^{1}-7)}$

$=\frac{48+16 \times \frac{1}{4}}{20-7}$

$=\frac{48+4}{13}$

$=\frac{52}{13}$

$=4$

Therefore, $\frac{6(8)^{n+1}+16(2)^{3 n-2}}{10(2)^{3 n+1}-7(8)^{n}}=4$. 

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Updated on: 10-Oct-2022

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