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by jordanriner26
Sun Oct 07, 2012 5:57 pm
Forum: Advanced Algebra ("pre-calculus")
Topic: expanding logs: solve [3e^{-5x}=132] algebraically
Replies: 1
Views: 1888

expanding logs: solve [3e^{-5x}=132] algebraically

Solve the equation algebraically. Approximate your result to three decimal places.

[3e^{-5x}=132]

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