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(3*2^(n)-4*2^(n-2))/(2^(n)-2^(n-1))

$\frac{3 \cdot 2^{n}-4 \cdot 2^{n-2}}{2^{n}-2^{n-1}}$

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Divide numbers written in scientific notation

Full solution

Q.

\frac{3 \cdot 2^{n}-4 \cdot 2^{n-2}}{2^{n}-2^{n-1}}

Simplify Numerator: First, let's simplify the numerator by factoring out the common term $2^{n-2}$ . $\newline$ $3\cdot2^n - 4\cdot2^{n-2} = 2^{n-2} \cdot (3\cdot2^2 - 4)$
Calculate Numerator: Now, calculate the value inside the parentheses. $\newline$ $3 \times 2^2 - 4 = 3 \times 4 - 4 = 12 - 4 = 8$ $\newline$ So, the numerator becomes $2^{(n-2)} \times 8$ .
Simplify Denominator: Next, let's simplify the denominator by factoring out the common term $2^{n-1}$ . $\newline$ $2^n - 2^{n-1} = 2^{n-1} \times (2 - 1)$
Calculate Denominator: Now, calculate the value inside the parentheses. $\newline$ $2 - 1 = 1$ $\newline$ So, the denominator becomes $2^{(n-1)} \times 1$ , which is simply $2^{(n-1)}$ .
Final Numerator and Denominator: Now we have the simplified numerator and denominator: $\newline$ Numerator: $2^{(n-2)} \times 8$ $\newline$ Denominator: $2^{(n-1)}$ $\newline$ We can now divide the numerator by the denominator. $\newline$ $(2^{(n-2)} \times 8) / (2^{(n-1)})$
Apply Exponent Rule: Apply the rule of exponents for division to simplify the expression. $2^{n-2} / 2^{n-1} = 2^{n-2-(n-1)} = 2^{-1} = \frac{1}{2}$
Multiply by Numerator: Now, multiply the result by $8$ (from the numerator). $(\frac{1}{2}) \times 8 = 4$
Final Simplified Form: The final simplified form of the expression is $4$ .

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