Question
(a) \( \frac{2^{x+2}-2^{x}}{2^{x}} \) (d) \( \frac{3^{m}+3^{m-1}}{2 \cdot 3^{m}} \) (g) \( \frac{2^{x}+2^{x+1}-2^{x+2}}{2^{x-3}} \)
Ask by Ramos Sandoval. in South Africa
Mar 09,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
**(a)**
\[
\frac{2^{x+2} - 2^x}{2^x} = 3
\]
**(d)**
\[
\frac{3^m + 3^{m-1}}{2 \cdot 3^m} = \frac{2}{3}
\]
**(g)**
\[
\frac{2^x + 2^{x+1} - 2^{x+2}}{2^{x-3}} = -8
\]
Solution
**(a)**
Rewrite the numerator by factoring \(2^x\):
\[
2^{x+2} - 2^x = 2^x \left(2^2 - 1\right) = 2^x (4 - 1) = 2^x \cdot 3.
\]
Then, the full expression becomes:
\[
\frac{2^x \cdot 3}{2^x} = 3.
\]
**(d)**
Rewrite the numerator by factoring \(3^{m-1}\):
\[
3^m + 3^{m-1} = 3^{m-1}\left(3 + 1\right) = 3^{m-1} \cdot 4.
\]
The expression is now:
\[
\frac{4 \cdot 3^{m-1}}{2 \cdot 3^m}.
\]
Since \(3^m = 3^{m-1} \cdot 3\), cancel \(3^{m-1}\) in the numerator and denominator:
\[
\frac{4}{2 \cdot 3} = \frac{4}{6} = \frac{2}{3}.
\]
**(g)**
Factor the numerator by extracting \(2^x\):
\[
2^x + 2^{x+1} - 2^{x+2} = 2^x \left(1 + 2 - 4\right) = 2^x \cdot (-1) = -2^x.
\]
Thus, the expression becomes:
\[
\frac{-2^x}{2^{x-3}}.
\]
Using the exponent rule \(\frac{2^x}{2^{x-3}} = 2^{x-(x-3)} = 2^3\), we have:
\[
-2^3 = -8.
\]
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Simplify this solution Extra Insights
Let’s dive into the fun world of exponents and explore how these expressions can simplify our calculations. For the first expression, \( \frac{2^{x+2}-2^{x}}{2^{x}} \), you can factor out \( 2^{x} \) from the numerator, giving you \( \frac{2^{x}(2^{2}-1)}{2^{x}} = 2^{2} - 1 = 4 - 1 = 3 \). Voila! It’s super simple when you break it down! Now, in the second expression, \( \frac{3^{m}+3^{m-1}}{2 \cdot 3^{m}} \), you can factor out \( 3^{m-1} \) giving \( \frac{3^{m-1}(3+1)}{2 \cdot 3^{m}} = \frac{4}{2 \cdot 3} = \frac{2}{3} \). Nice and tidy, isn't it? By applying these clever factoring techniques, simplifying expressions can feel almost like magic!
