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Home » $\boxed{\dfrac10x - 43}$ - Aurero

$\boxed{\dfrac10x - 43}$ - Aurero

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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Thus, \(r^2 = 24\) and \(r = \sqrt{24} = 2\sqrt{6}\).
#### \(2\sqrt{6}\)
Solve for \(x\): \(\log_2(x^2 - 5x) = 3\).
Rewrite: \(x^2 - 5x = 2^3 = 8\).
So, \(x^2 - 5x - 8 = 0\).
Use quadratic formula: \(x = \frac{5 \pm \sqrt{25 + 32}}{2}\).
\(x = \frac{5 \pm \sqrt{57}}{2}\).
Approximately, \(x \approx 6.27\) or \(x \approx -1.27\).
Only \(x \approx 6.27\) is valid since \(x^2 - 5x > 0\).
#### 6.27Question: Let $ x, y, z $ be positive real numbers such that...

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