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1) \( a^{6} \cdot a^{3}= \) 2) \( a^{-5} \cdot a= \) 3) \( b \cdot b^{x}= \) 4) \( 2^{5} \cdot 2^{2}= \) 5) \( (P s)^{6}= \) 6) \( \left(b^{-2}\right)^{-8}= \) 7) \( (-3)^{a}-4^{a}= \) 8) \( \left(\frac{1}{3}\right)^{x} \cdot\left(\frac{6}{5}\right)^{x} \) a) \( (3 x)^{2}= \) (0) \( \left(3 m n^{2}\right)^{4}= \) 4) \( \left[\frac{a^{2}}{a^{3}}\right)^{-3} \) 12) \( \left[3 \times Y 7^{2}\right]^{0} \)

Ask by Wheeler Cummings. in Colombia
Feb 26,2025

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Answer

1) \( a^{9} \) 2) \( a^{-4} \) 3) \( b^{x+1} \) 4) \( 2^{7} \) 5) \( P^{6} \cdot s^{6} \) 6) \( b^{16} \) 7) \( (–3)^{a} – 4^{a} \) 8) \( \left(\frac{2}{5}\right)^{x} \) 9) \( 9x^{2} \) 10) \( 81m^{4}n^{8} \) 11) \( a^{3} \) 12) \( 1 \)

Solution

Aquí tienes las soluciones paso a paso: 1) a⁶ · a³   Usamos la propiedad de los exponentes: aᵐ · aⁿ = aᵐ⁺ⁿ.   Resultado: a^(6+3) = a⁹. 2) a⁻⁵ · a   Observamos que a = a¹, por lo que: a⁻⁵ · a¹ = a^(–5+1) = a⁻⁴. 3) b · bˣ   Aplicamos la misma regla: b¹ · bˣ = b^(1+x) = b^(x+1). 4) 2⁵ · 2²   Sumamos los exponentes: 2⁵ · 2² = 2^(5+2) = 2⁷. 5) (P s)⁶   Se eleva cada factor a la sexta potencia: (P s)⁶ = P⁶ · s⁶. 6) (b⁻²)⁻⁸   Multiplicamos los exponentes: (b⁻²)⁻⁸ = b^(–2·–8) = b¹⁶. 7) (–3)ᵃ – 4ᵃ   No se pueden simplificar más ya que los términos tienen bases diferentes.   Resultado: (–3)ᵃ – 4ᵃ. 8) (1/3)ˣ · (6/5)ˣ   Factorizamos usando la propiedad de exponentes: (1/3 · 6/5)ˣ = (6/15)ˣ.   Como 6/15 se simplifica a 2/5, queda: (2/5)ˣ. 9) (3x)²   Se eleva cada factor: (3x)² = 3² · x² = 9x². 10) (3 m n²)⁴   Aplicamos la propiedad a cada factor: 3⁴ · m⁴ · (n²)⁴ = 81 · m⁴ · n^(2·4) = 81 m⁴ n⁸. 11) [(a²)/(a³)]⁻³   Primero simplificamos el cociente: a² / a³ = a^(2–3) = a⁻¹.   Luego, (a⁻¹)⁻³ = a^(–1·–3) = a³. 12) [3 × Y 7²]⁰   Cualquier número (distinto de cero) elevado a la potencia cero es 1.   Resultado: 1. Espero que estas soluciones te sean de ayuda.

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Mind Expander

When you multiply terms with the same base, you simply add their exponents! This principle is grounded in the rules of exponents, making calculations with powers easier and more efficient. For instance, \( a^6 \cdot a^3 \) becomes \( a^{6+3} = a^9 \). Simple, right? Don't forget about the special cases like any non-zero number raised to the zero power equals one! That’s why for \( \left[3 \times Y 7^{2}\right]^{0} \), regardless of the values of \( Y \) or \( 7 \), the result is simply 1. Always a fun surprise when you run into those unique exponent properties!

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