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\( \log _ { 5 } ^ { x ^ { 2 } + 3 x } \)

Ask by Bryant Rodriquez. in Angola
Jan 23,2025

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Answer

A expressão \( \log_{5}(x^{2} + 3x) \) calcula a potência a que a base 5 deve ser elevada para obter o valor de \( x^{2} + 3x \).

Solution

A expressão \( \log_{5}(x^{2} + 3x) \) representa o logaritmo de \( x^{2} + 3x \) na base 5. Em outras palavras, ela responde à seguinte pergunta: **A que potência a base 5 deve ser elevada para resultar em \( x^{2} + 3x \)?** Matematicamente, isso significa: \[ 5^{\log_{5}(x^{2} + 3x)} = x^{2} + 3x \] **Interpretação:** - **Base do Logaritmo:** 5 - **Argumento do Logaritmo:** \( x^{2} + 3x \) - **Resultado do Logaritmo:** O expoente ao qual a base 5 deve ser elevada para obter \( x^{2} + 3x \). **Exemplo de Aplicação:** Suponha que \( x = 2 \): 1. Calcule o argumento do logaritmo: \[ x^{2} + 3x = 2^{2} + 3(2) = 4 + 6 = 10 \] 2. Encontre \( \log_{5}(10) \), que é o valor de \( y \) tal que \( 5^{y} = 10 \). Esse tipo de logaritmo é útil em diversas áreas, como resolução de equações exponenciais, modelagem de crescimento e decaimento, e em escalas logarítmicas utilizadas em ciência e engenharia.

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Did you know that logarithms have their roots in the work of John Napier, who introduced them in the early 17th century to simplify complex calculations? His invention was a game-changer for astronomers and navigators who relied on accurate calculations for their work. Logarithms turned multiplication and division into easier addition and subtraction tasks! When working with logarithms, a common mistake is forgetting the change of base formula. If you ever need to convert \( \log_b(a) \) to a more manageable base (like 10 or \( e \)), you can use \( \log_b(a) = \frac{\log_k(a)}{\log_k(b)} \) where \( k \) can be any positive number. This makes calculations much smoother and opens up new realms of understanding for logarithmic functions!

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