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Ch. 4 - Exponential and Logarithmic Functions
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Non è quello che usi tu?Cambia libro di testo
Capitolo 5, Problema 101

Solve each equation. 5x2−12=252x5^{x^2 - 12} = 25^{2x}

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1
Recognize that both sides of the equation involve exponential expressions with base 5 or powers of 5. Rewrite 25 as a power of 5: since \(25 = 5^2\), rewrite the right side as \(25^{2x} = (5^2)^{2x}\).
Apply the power of a power property: \((a^m)^n = a^{mn}\). So, \((5^2)^{2x} = 5^{4x}\). Now the equation becomes \(5^{x^2 - 12} = 5^{4x}\).
Since the bases are the same and the expressions are equal, set the exponents equal to each other: \(x^2 - 12 = 4x\).
Rewrite the equation to standard quadratic form by moving all terms to one side: \(x^2 - 4x - 12 = 0\).
Solve the quadratic equation \(x^2 - 4x - 12 = 0\) using factoring, completing the square, or the quadratic formula to find the values of \(x\).

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