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📅 May 15, 2026 👤 scraface

May 15, 2026

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🔥 Popular Posts

Start by rewriting the logarithmic equation in exponential form:
\[ \log_2(x^2 - 4) = 3 \Rightarrow x^2 - 4 = 2^3 = 8. \]
Solve for \( x^2 \):
\[ x^2 = 8 + 4 = 12 \Rightarrow x = \pm \sqrt{12} = \pm 2\sqrt{3}. \]
Since the argument of a logarithm must be positive and \( x^2 - 4 = 8...
Thus, one solution is \( \boxed{2\sqrt{3}} \).
Clarification Note:** Since \( \log_2(x^2 - 4) = 3 \) requires \( x^2...
\boxed{2\sqrt{3}}
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📝 Recent Posts

So \( x = 0 \) is the only solution — but this implies the circle is t...
At \( x = 0 \): \( y^2 = 64 \), so \( y = \pm8 \).
Check in ellipse: \( 0 + \frac{64}{64} = 1 \) — yes.
Now check if other intersections exist:
The circle has radius 8, ellipse semi-minor axis is 8 — so the circle...
But does it intersect in more points? No — since the circle lies entir...
Actually, the circle is inscribed: at \( x=0 \), both have \( y = \pm8...
Try solving numerically:
At \( x = 8 \), ellipse: \( y^2 = 64(1 - 64/100) = 64(0.36) = 23.04 \)...
Circle: at \( x=8 \), \( y=0 \), so not intersecting.

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