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

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
jordanriner26
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expanding logs: solve [3e^{-5x}=132] algebraically

Postby jordanriner26 » Sun Oct 07, 2012 5:57 pm

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

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

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

Postby maggiemagnet » Sun Oct 07, 2012 11:05 pm

jordanriner26 wrote:Solve the equation algebraically. Approximate your result to three decimal places.

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

Start by dividing off the 3. Then log both sides and divide by the -5. You can see more here.
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