Simplify each expression as far as possible: (a) \(\mathrm{e}^{2 x} \mathrm{e}^{7 x}\) (b) \(\left(3 \mathrm{e}^{x}\right)\left(2 \mathrm{e}^{-x}\right)\) (c) \(\mathrm{e}^{2 x}\left(\mathrm{e}^{-2 x}+\mathrm{e}^{-x}+1\right)-\mathrm{e}^{x}\left(1+\mathrm{e}^{x}\right)\) (d) \(\frac{\mathrm{e}^{-3 x}}{2 \mathrm{e}^{-x}}\)

Short Answer

Expert verified
Short Answer: The simplified expressions are: (a) \(\mathrm{e}^{9x}\), (b) \(6\), (c) \(1\), and (d) \(\frac{1}{2}\mathrm{e}^{-2x}\).

Step by step solution

01

Expression (a) Simplification

Given the expression \(\mathrm{e}^{2 x} \mathrm{e}^{7 x}\), apply the properties of exponents: \(\mathrm{e}^{2x} \cdot \mathrm{e}^{7x} = \mathrm{e}^{2x + 7x} = \boxed{\mathrm{e}^{9x}}\).
02

Expression (b) Simplification

For the expression \(\left(3 \mathrm{e}^{x}\right)\left(2 \mathrm{e}^{-x}\right)\), apply the properties of exponents and multiplication: \((3 \mathrm{e}^{x})(2 \mathrm{e}^{-x}) = 3 \cdot 2 \cdot \mathrm{e}^{x} \cdot \mathrm{e}^{-x} = 6 \cdot \mathrm{e}^{x-x} = \boxed{6}\).
03

Expression (c) Simplification

Simplify the expression \(\mathrm{e}^{2x}\left(\mathrm{e}^{-2x}+\mathrm{e}^{-x}+1\right)-\mathrm{e}^{x}\left(1+\mathrm{e}^{x}\right)\) as follows: 1. Distribute \(\mathrm{e}^{2x}\) across the terms within the parentheses: \(\mathrm{e}^{2x}\mathrm{e}^{-2x}+\mathrm{e}^{2x}\mathrm{e}^{-x}+\mathrm{e}^{2x}-\mathrm{e}^{x}\left(1+\mathrm{e}^{x}\right)\) 2. Apply the properties of exponents: \(1+\mathrm{e}^{x}+\mathrm{e}^{2x}-\mathrm{e}^{x}\left(1+\mathrm{e}^{x}\right)\) 3. Distribute \(-\mathrm{e}^{x}\) across the terms within the parentheses: \(1+\mathrm{e}^{x}+\mathrm{e}^{2x}-\mathrm{e}^{x}-\mathrm{e}^{2x}\) 4. Combine like terms: \(1 + \left(\mathrm{e}^{x} - \mathrm{e}^{x}\right) + \left(\mathrm{e}^{2x} - \mathrm{e}^{2x}\right) = \boxed{1}\).
04

Expression (d) Simplification

Finally, simplify the expression \(\frac{\mathrm{e}^{-3x}}{2\mathrm{e}^{-x}}\): 1. Apply the properties of exponents: \(\frac{\mathrm{e}^{-3x}}{2\mathrm{e}^{-x}} = \frac{1}{2} \cdot \frac{\mathrm{e}^{-3x}}{\mathrm{e}^{-x}}\). 2. Continue applying the properties of exponents: \(\frac{1}{2} \cdot \mathrm{e}^{-3x + x} = \boxed{\frac{1}{2}\mathrm{e}^{-2x}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free