Example Question - exponents properties
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Algebraic Expression Simplification
Para resolver la pregunta proporcionada en la imagen, parece que tenemos que simplificar la expresión algebraica. La expresión es: \[\frac{x^2 y^{-1/3}}{x^{-3} y^{1/2}} - \frac{7x^2}{2xy} \cdot \frac{10x^{-1}y}{x^{-1/2}y^1} + \frac{4x^5}{12x^4y}\] Empecemos por simplificar cada término individualmente: Para el primer término, utilizamos las propiedades de los exponentes para dividir las bases iguales restando los exponentes: \[\frac{x^2 y^{-1/3}}{x^{-3} y^{1/2}} = x^{2 - (-3)} y^{-\frac{1}{3} - \frac{1}{2}} = x^{2 + 3} y^{-\frac{1}{3} - \frac{2}{6}} = x^5 y^{-\frac{3}{6}} = x^5 y^{-\frac{1}{2}}\] Para el segundo término, primero simplificamos la fracción compuesta: \[\frac{7x^2}{2xy} \cdot \frac{10x^{-1}y}{x^{-1/2}y^1} = \frac{7x^2 \cdot 10x^{-1}y}{2xy \cdot x^{-1/2}y} = \frac{70x^{2-1}}{2x^{1-(-1/2)}y^{1+1}} = \frac{70x^{1}}{2x^{3/2}y^{2}}\] Dado que \(70/2 = 35\) y \(x^{1 - 3/2} = x^{-1/2}\), tenemos: \[\frac{70x^{1}}{2x^{3/2}y^{2}} = \frac{35}{y^{2}x^{1/2}}\] Para el tercer término, simplemente reducimos la fracción \(\frac{4x^5}{12x^4y}\): \[\frac{4x^5}{12x^4y} = \frac{4}{12} \cdot \frac{x^5}{x^4y} = \frac{1}{3} x^{5-4}y^{-1} = \frac{1}{3}xy^{-1}\] Ahora combinamos los términos simplificados: \[x^5 y^{-\frac{1}{2}} - \frac{35}{y^2x^{1/2}} + \frac{1}{3}xy^{-1}\] Este es la expresión simplificada de términos individuales de la expresión inicial en la imagen. Dada la complejidad de combinarlos en una expresión simple y común, podría ser más apropiado dejarlos en esta forma, a menos que tengamos un denominador común o una indicación de que necesitamos combinarlos de alguna manera específica. Si se necesita una simplificación adicional o una manipulación adicional, eso requeriría un contexto adicional o instrucciones específicas.
Solving Exponential Expressions
To solve the given expression, we will use the properties of exponents: The expression is \( \left( \frac{(5^3)^2}{(5^2)^4} \right) \). Now, let's apply the power of a power property for exponents, which states that \((a^m)^n = a^{(m \cdot n)}\): For the numerator, \((5^3)^2 = 5^{(3 \cdot 2)} = 5^6\), For the denominator, \((5^2)^4 = 5^{(2 \cdot 4)} = 5^8\), Now, rewrite the expression with the simplified exponents: \( \frac{5^6}{5^8} \). Using the quotient of powers property, which states that \(a^m/a^n = a^{(m-n)}\), we subtract the exponents: \( 5^{(6-8)} = 5^{-2} \). The negative exponent rule states that \(a^{-n} = \frac{1}{a^n}\): Therefore, \(5^{-2} = \frac{1}{5^2}\). Since \(5^2 = 25\), the expression simplifies to: \( \frac{1}{25} \). Therefore, the answer is option (1) \( \frac{1}{25} \).
Simplifying Exponential Expressions
To simplify the expression \(\frac{2^{n+1}}{2^n * 2}\), you can use the properties of exponents. Here is how you can simplify it step by step: 1. The expression \(\frac{2^{n+1}}{2^n * 2}\) can be rewritten by separating the terms in the denominator, so you have \(\frac{2^{n+1}}{2^n \cdot 2^1}\). 2. Now you can apply the quotient rule for exponents, which states that when dividing like bases, you subtract the exponents: \(\frac{2^{n+1}}{2^n}\) becomes \(2^{(n+1)-n} = 2^1 = 2\). 3. So now you have \(2 \div 2^1\). 4. Applying the division rule for exponents again (or recognizing that any number divided by itself is 1), you have \(2^1 \div 2^1 = 2^{1-1} = 2^0 = 1\). The simplified form of the expression is \(1\).
Solving a System of Exponential Equations
The image shows a system of equations that needs to be solved: \(2^{x+y} = 8\) and \(2^{3x-y} = 128\) For the first equation, we can rewrite 8 as \(2^3\) since \(2^3 = 8\). This gives: \(2^{x+y} = 2^3\) From the property of exponents that states if \(a^m = a^n\) then \(m=n\) for any non-zero base \(a\), we get: \(x + y = 3\) (Equation 1) For the second equation, we can rewrite 128 as \(2^7\), since \(2^7 = 128\). This gives: \(2^{3x-y} = 2^7\) Using the same property of exponents, we get: \(3x - y = 7\) (Equation 2) Now we have a system of two linear equations in two variables (\(x\) and \(y\)): 1. \(x + y = 3\) 2. \(3x - y = 7\) We can solve this system by adding Equation 1 and Equation 2 together to eliminate \(y\): \(x + y + 3x - y = 3 + 7\) \(4x = 10\) \(x = 10/4\) \(x = 2.5\) With \(x\) found, we can substitute it back into Equation 1 to find \(y\): \(2.5 + y = 3\) \(y = 3 - 2.5\) \(y = 0.5\) Therefore, the solution to the system is \(x = 2.5\) and \(y = 0.5\).
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