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Simplify the following expressions involving exponents: a) (3^(n+1) + 9 3ⁿ) / (6 3ⁿ + 3^(n+1)) b) (5^(-2x+1) - 25^x) / (3 * 5^(-2x+1) + 25^(-x))

Mathematics

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Answer

a

43\frac{4}{3}

Step-by-step solution

8 steps

Here are the solutions for the simplification problems.

Part a· 3 steps

Simplify 3n+1+9⋅3n6⋅3n+3n+1\frac{3^{n+1} + 9 \cdot 3^n}{6 \cdot 3^n + 3^{n+1}}

  1. Rewrite terms using exponent rules (am+n=am⋅ana^{m+n} = a^m \cdot a^n) and express 9 as a power of 3.

    3n+1=3n⋅31=3⋅3n3^{n+1} = 3^n \cdot 3^1 = 3 \cdot 3^n 9=329 = 3^2

    Substitute these into the expression:

    3⋅3n+32⋅3n6⋅3n+3⋅3n\frac{3 \cdot 3^n + 3^2 \cdot 3^n}{6 \cdot 3^n + 3 \cdot 3^n}

  2. Factor out the common term 3n3^n from both the numerator and the denominator.

    3n(3+32)3n(6+3)\frac{3^n (3 + 3^2)}{3^n (6 + 3)} 3n(3+9)3n(9)\frac{3^n (3 + 9)}{3^n (9)} 3n(12)3n(9)\frac{3^n (12)}{3^n (9)}

  3. Cancel out 3n3^n and simplify the fraction.

    129=43\frac{12}{9} = \frac{4}{3}

    The simplified expression is 43\boxed{\frac{4}{3}}.

Part b· 5 steps

Simplify 5−2x+1−25x3⋅5−2x+1+25−x\frac{5^{-2x+1} - 25^x}{3 \cdot 5^{-2x+1} + 25^{-x}}

  1. Rewrite all terms with base 5 using exponent rules (am+n=am⋅ana^{m+n} = a^m \cdot a^n and (am)n=amn(a^m)^n = a^{mn}).

    5−2x+1=5−2x⋅51=5⋅5−2x5^{-2x+1} = 5^{-2x} \cdot 5^1 = 5 \cdot 5^{-2x} 25x=(52)x=52x25^x = (5^2)^x = 5^{2x} 25−x=(52)−x=5−2x25^{-x} = (5^2)^{-x} = 5^{-2x}

  2. Substitute these rewritten terms into the expression.

    Numerator:

    5−2x+1−25x=5⋅5−2x−52x5^{-2x+1} - 25^x = 5 \cdot 5^{-2x} - 5^{2x}

    Denominator:

    3⋅5−2x+1+25−x=3⋅(5⋅5−2x)+5−2x3 \cdot 5^{-2x+1} + 25^{-x} = 3 \cdot (5 \cdot 5^{-2x}) + 5^{-2x} =15⋅5−2x+5−2x= 15 \cdot 5^{-2x} + 5^{-2x}

    Factor out 5−2x5^{-2x} from the denominator:

    =5−2x(15+1)=16⋅5−2x= 5^{-2x} (15 + 1) = 16 \cdot 5^{-2x}

  3. Form the simplified fraction.

    5⋅5−2x−52x16⋅5−2x\frac{5 \cdot 5^{-2x} - 5^{2x}}{16 \cdot 5^{-2x}}

  4. Divide each term in the numerator by the denominator.

    5⋅5−2x16⋅5−2x−52x16⋅5−2x\frac{5 \cdot 5^{-2x}}{16 \cdot 5^{-2x}} - \frac{5^{2x}}{16 \cdot 5^{-2x}}

  5. Simplify each term.

    The first term:

    5⋅5−2x16⋅5−2x=516\frac{5 \cdot 5^{-2x}}{16 \cdot 5^{-2x}} = \frac{5}{16}

    The second term:

    52x16⋅5−2x=116⋅52x5−2x\frac{5^{2x}}{16 \cdot 5^{-2x}} = \frac{1}{16} \cdot \frac{5^{2x}}{5^{-2x}}

    Using the exponent rule $\frac{a^m}{a^n} = a

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