$\frac{2^{n+4} - 2(2^n)}{2(2^{n+3})}$ when simplified is

### Concept & Formula When simplifying algebraic fractions with variable exponents, the goal is to factor out the smallest common power of the base from both the numerator and the denominator so they can be canceled out. The key laws of indices required are: $$x^m \times x^n = x^{m+n}$$ $$x^{m+n} = x^m \times x^n$$ ### Step-by-Step Solution * **Given:** $$\frac{2^{n+4} - 2(2^n)}{2(2^{n+3})}$$ * **Calculation:** First, break down the exponents in the numerator using the product rule in reverse ($x^{m+n} = x^m \cdot x^n$): Numerator $= 2^n \cdot 2^4 - 2^1 \cdot 2^n$ Factor out the common term, $2^n$, from the numerator: $$2^n(2^4 - 2^1) = 2^n(16 - 2) = 2^n \cdot 14$$ Now, simplify the denominator. Remember that $2$ is $2^1$: Denominator $= 2^1 \cdot 2^{n+3} = 2^{1 + n + 3} = 2^{n+4}$ Break this down to isolate $2^n$: $$2^{n+4} = 2^n \cdot 2^4 = 2^n \cdot 16$$ Substitute the simplified numerator and denominator back into the fraction: $$\frac{2^n \cdot 14}{2^n \cdot 16}$$ Cancel out the common $2^n$ term from the top and bottom: $$\frac{14}{16}$$ Reduce the fraction to its lowest terms by dividing by $2$: $$\frac{7}{8}$$ ### Exam Strategy & Shortcut **Value Substitution (Plug-in Method):** Since the options primarily consist of constants (and one of them is the correct answer), this means the expression must hold true for any arbitrary value of $n$. Assume $n = 0$ for the easiest calculation: Numerator: $2^{(0+4)} - 2(2^0) = 2^4 - 2(1) = 16 - 2 = 14$ Denominator: $2(2^{(0+3)}) = 2(2^3) = 2(8) = 16$ Result: $\frac{14}{16} = \frac{7}{8}$. This perfectly matches option (d) in just a few seconds of basic arithmetic. ### Common Pitfall A standard error occurs when students try to subtract exponents across subtraction signs in the numerator, incorrectly attempting to apply division rules to individual terms before factoring. Always factor completely before attempting to cancel terms across a division line. ### Final Answer **Therefore, the correct answer is $\frac{7}{8}$.**