# How to solve (2 ^ (x + 4) – 2 ^ 5 * 2 ^ x)/(2 * 2 ^ (x + 3)) – 2 ^ – 3

Welcome to my article How to solve (2 ^ (x + 4) – 2 ^ 5 * 2 ^ x)/(2 * 2 ^ (x + 3)) – 2 ^ – 3. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve (2 ^ (x + 4) – 2 ^ 5 * 2 ^ x)/(2 * 2 ^ (x + 3)) – 2 ^ – 3, read and understand it carefully till the end.

Let us know how to solve this question How to solve (2 ^ (x + 4) – 2 ^ 5 * 2 ^ x)/(2 * 2 ^ (x + 3)) – 2 ^ – 3.

First write the question on the page of the notebook.

## How to solve (2 ^ (x + 4) – 2 ^ 5 * 2 ^ x)/(2 * 2 ^ (x + 3)) – 2 ^ – 3

To solve this question we will write in this way,

\displaystyle \frac{{({{2}^{{(x+4)}}}-{{2}^{5}}\times {{2}^{x}})}}{{(2\times {{2}^{{(x+3)}}})}}-1\times {{2}^{{-3}}}

\displaystyle \frac{{({{2}^{{(x+4)}}}-32\times {{2}^{x}})}}{{({{2}^{1}}\times {{2}^{{(x+3)}}})}}-1\times {{2}^{{-3}}}

\displaystyle \frac{{({{2}^{{(x+4)}}}-32\times {{2}^{x}})}}{{({{2}^{{(x+3)+1}}})}}-1\times {{2}^{{-3}}}

\displaystyle \frac{{({{2}^{{(x+4)}}}-32\times {{2}^{x}})}}{{({{2}^{{(x+3+1)}}})}}-1\times \frac{1}{{{{2}^{3}}}}

\displaystyle \frac{{({{2}^{{(x+4)}}}-32\times {{2}^{x}})}}{{({{2}^{{(x+4)}}})}}-1\times \frac{1}{8}

\displaystyle \frac{{({{2}^{{(x+4)}}}-{{2}^{1}}\times {{2}^{4}}\times {{2}^{x}})}}{{({{2}^{{(x+4)}}})}}-\frac{1}{8}

\displaystyle \frac{{({{2}^{{(x+4)}}}-2\times {{2}^{{(x+4)}}})}}{{({{2}^{{(x+4)}}})}}-\frac{1}{8}

\displaystyle \frac{{(1\times {{2}^{{(x+4)}}}-2\times {{2}^{{(x+4)}}})}}{{1\times ({{2}^{{(x+4)}}})}}-\frac{1}{8}

\displaystyle \frac{{({{2}^{{(x+4)}}})}}{{({{2}^{{(x+4)}}})}}\left[ {\frac{{(1-2)}}{1}} \right]-\frac{1}{8}

\displaystyle \frac{{\cancel{{({{2}^{{(x+4)}}})}}}}{{\cancel{{({{2}^{{(x+4)}}})}}}}\left[ {\frac{{(1-2)}}{1}} \right]-\frac{1}{8}

\displaystyle \left[ {\frac{{-1}}{1}} \right]-\frac{1}{8}

\displaystyle \frac{{-8-1}}{8}

\displaystyle \frac{{-9}}{8}