Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 5: Problem 93
Solve using any method. $$\frac{\sqrt{\left(e^{2 x} \cdot e^{-5 x}\right)^{-4}}}{e^{x} \div e^{-x}}=e^{7}$$
Short Answer
Expert verified
The value of \(x\) is \(-\frac{1}{2}\).
Step by step solution
01
- Simplify the expression inside the square root
Combine the exponents inside the exponential terms: \[ e^{2x} \times e^{-5x} = e^{2x - 5x} = e^{-3x} \]
02
- Apply the exponent outside the square root
Apply the exponent \(-4\) to the simplified term: \[ (e^{-3x})^{-4} = e^{12x} \]
03
- Simplify the numerator
Apply the square root to the term: \[ \frac{\textcolor{blue}{\text{numerator}}}{\text{denominator}} = \frac{\textcolor{blue}{\text{simplified term}}}{\text{denominator}} \rightarrow \frac{\textcolor{blue}{\frac{1}{e^{12x}}}}{e^{x} \frac{e^{x}}{e^{-x}}}\] Consider that: \[ \frac{1}{e^{12x}} = e^{-12x} \]
04
- Simplify the denominator
Combine the terms in the denominator: \[ e^{x} \times \frac{1}{e^{-x}} = e^{x} \times e^{x} = e^{2x} \]
05
- Combine the simplified terms
Substitute the simplified forms back into the original equation: \[ \frac{e^{-12x}}{e^{2x}} = e^{-12x - 2x} = e^{-14x} \]
06
- Set the equation equal to \(e^7\)
Set the simplified form equal to the given expression: \[ e^{-14x} = e^7 \]
07
- Solving for \(x\)
Since the bases are the same, set the exponents equal to each other: \[ -14x = 7 \] Solve for \(x\): \[ x = -\frac{7}{14} = -\frac{1}{2} \]
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 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
exponents
Exponents are a fundamental concept in mathematics that involve numbers raised to a power. This is expressed as \(a^b\), where \(a\) is the base and \(b\) is the exponent. In this exercise, we work with the exponential function \(e^x\), where \(e\) is the base and \(x\) is the exponent. When handling exponents, some key rules to remember are:
- Multiplication of same bases: \(a^m \times a^n = a^{m+n}\)
- Division of same bases: \(\frac{a^m}{a^n} = a^{m-n}\)
- Power of a power: \((a^m)^n = a^{m \times n}\)
- Negative exponent: \(a^{-n} = \frac{1}{a^n}\)
simplification
Simplification involves reducing expressions to their simplest forms, making them easier to work with. Let's break down our steps.In Step 1, we combine exponents within a product: \(e^{2x} \times e^{-5x} = e^{2x - 5x} = e^{-3x}\). Next, we apply the given exponent to this term in Step 2: \((e^{-3x})^{-4} = e^{-3x \times -4} = e^{12x}\). Step 4 simplifies the denominator: \(e^{x} \times \frac{1}{e^{-x}} = e^{x} \times e^{x} = e^{2x}\). The entire numerator becomes: \(\frac{1}{e^{12x}} = e^{-12x}\). These steps streamline the original complex equation to forms that are easier to handle.
solving for x
Solving for \(x\) means isolating \(x\) on one side of the equation. After simplification, we have: \( \frac{e^{-12x}}{e^{2x}} = e^{7}\). Combining terms in the numerator and denominator gets us: \(e^{-12x} \times e^{-2x} = e^{-14x}\). Setting the simplified expression equal to the provided value: \(e^{-14x} = e^{7}\). Since the exponential function with the same base \(e\) is equal only if their exponents are equal, we set \(-14x = 7\). Solving this linear equation gives us: \(x = -\frac{1}{2}\).
right triangle approach
The right triangle approach is a general problem-solving strategy that can also apply to simplifying exponential terms. Imagine the ratio of the sides of a right triangle analogous to the steps in our simplification process.
- Determine one leg of the triangle (inequivalent but useful step such as combining like terms).
- Find the hypotenuse (main objective like finding usable simplified expressions).
- Solve for the other leg (solving for x with simpler terms we achieved).
