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

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

Postby jordanriner26 on 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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Re: expanding logs: solve [3e^{-5x}=132] algebraically  TOPIC_SOLVED

Postby maggiemagnet on 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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