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

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

- Sun Oct 07, 2012 5:57 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: expanding logs: solve [3e^{-5x}=132] algebraically
- Replies:
**1** - Views:
**2298**

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

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

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