Simplify the following: (2^(n+5) - 3 * 2^(n-2))/(5 * 2^(n+3))

Steps to Solve

1. Rewrite the numerator using exponent rules

We will rewrite $2^{n+5}$ as $2^n \cdot 2^5$ and $2^{n-2}$ as $2^n \cdot 2^{-2}$. This allows us to factor out $2^n$ from the numerator. $$ \frac{2^{n+5} - 3 \cdot 2^{n-2}}{5 \cdot 2^{n+3}} = \frac{2^n \cdot 2^5 - 3 \cdot 2^n \cdot 2^{-2}}{5 \cdot 2^{n+3}} $$

2. Factor out $2^n$ from the numerator

We can now factor $2^n$ from the numerator: $$ \frac{2^n (2^5 - 3 \cdot 2^{-2})}{5 \cdot 2^{n+3}} $$

3. Simplify the terms within the parenthesis

Let's simplify $2^5$ and $2^{-2}$. $2^5 = 32$ and $2^{-2} = \frac{1}{2^2} = \frac{1}{4}$. Substitute these back into the expression: $$ \frac{2^n (32 - 3 \cdot \frac{1}{4})}{5 \cdot 2^{n+3}} = \frac{2^n (32 - \frac{3}{4})}{5 \cdot 2^{n+3}} $$ Now, simplify the expression inside the parenthesis: $32 - \frac{3}{4} = \frac{32 \cdot 4 - 3}{4} = \frac{128 - 3}{4} = \frac{125}{4}$ So the expression becomes: $$ \frac{2^n \cdot \frac{125}{4}}{5 \cdot 2^{n+3}} $$

4. Rewrite the denominator using exponent rules Now we handle the denominator $5 \cdot 2^{n+3} = 5 \cdot 2^n \cdot 2^3 = 5 \cdot 2^n \cdot 8 = 40 \cdot 2^n$. The whole expression now looks like this: $$ \frac{2^n \cdot \frac{125}{4}}{40 \cdot 2^n} $$

5. Cancel out common terms

Cancel out $2^n$ from the numerator and the denominator: $$ \frac{\frac{125}{4}}{40} $$

6. Simplify the fraction

To simplify the fraction, we divide $\frac{125}{4}$ by $40$: $$ \frac{125}{4} \div 40 = \frac{125}{4} \cdot \frac{1}{40} = \frac{125}{4 \cdot 40} = \frac{125}{160} $$ Now, simplify the fraction by dividing both numerator and denominator by their greatest common divisor, which is 5: $$ \frac{125 \div 5}{160 \div 5} = \frac{25}{32} $$