Chapter 11: Problem 49
Explain how you would solve the equation \(2^{x}=64\) and also how you would solve the equation \(2^{x}=53\).
Short Answer
Expert verified
For \(2^x = 64\), \(x = 6\). For \(2^x = 53\), \(x \approx 5.73\).
Step by step solution
01
Recognize Powers for Integer Solutions
First, for the equation \(2^x = 64\), identify if 64 is a power of 2. We know that 64 can be expressed as \(2^6\) because \(2 \times 2 \times 2 \times 2 \times 2 \times 2 = 64\).
02
Solve for x in Integer Solution
Since we have \(2^x = 64\) and \(64 = 2^6\), we can conclude that \(x = 6\). This is because the bases are the same and thus the exponents must be equal.
03
Recognize Non-Integer Powers for Approximation
For \(2^x = 53\), recognize that 53 is not a power of 2. We need to use logarithms to find a non-integer solution.
04
Use Logarithms for Approximation
Take the logarithm of both sides: \(\log(2^x) = \log(53)\). Use the power rule of logarithms to rewrite as \(x \cdot \log(2) = \log(53)\).
05
Solve for x in Non-Integer Solution
Rearrange the equation to solve for \(x\): \(x = \frac{\log(53)}{\log(2)}\). Calculate this using a calculator: \(x \approx 5.73\).
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Powers of 2
Understanding powers of 2 is crucial when dealing with exponential equations like \(2^x = 64\). Powers of 2 are simply 2 multiplied by itself a number of times. The solutions to exponential equations that equate to a power of 2 can often be found by identifying these common powers.
For example, 64 is a known power of 2 because \(2^6 = 64\). We arrive at this by multiplying 2 by itself six times:
When approaching equations involving powers of 2, aim to recognize or convert numbers into powers of 2 to find solutions quickly and efficiently.
For example, 64 is a known power of 2 because \(2^6 = 64\). We arrive at this by multiplying 2 by itself six times:
- \(2 \times 2 = 4\)
- \(4 \times 2 = 8\)
- \(8 \times 2 = 16\)
- \(16 \times 2 = 32\)
- \(32 \times 2 = 64\)
When approaching equations involving powers of 2, aim to recognize or convert numbers into powers of 2 to find solutions quickly and efficiently.
Logarithms in Algebra
When the number on the right side of an equation like \(2^x = 53\) is not a power of 2, logarithms become a valuable tool. They allow us to find non-integer values for \(x\).
Solving \(x \cdot \log(2) = \log(53)\) involves rearranging to:
- Begin by taking the logarithm of both sides: \(\log(2^x) = \log(53)\)
- Use the power rule for logarithms: \(x \cdot \log(2) = \log(53)\)
Solving \(x \cdot \log(2) = \log(53)\) involves rearranging to:
- \(x = \frac{\log(53)}{\log(2)}\)
Approximation Methods
Approximation is sometimes necessary, especially when solving for non-integer solutions. In equations like \(2^x = 53\), approximation provides a practical solution to real-world problems.
Exact solutions may not always be possible or necessary. Even after using logarithms to solve \(x = \frac{\log(53)}{\log(2)}\), a calculator's approximation to \(x \approx 5.73\) suffices.
Here are some scenarios where approximation methods are useful:
Being adept at knowing when and how to approximate is a vital skill when tackling exponential equations.
Exact solutions may not always be possible or necessary. Even after using logarithms to solve \(x = \frac{\log(53)}{\log(2)}\), a calculator's approximation to \(x \approx 5.73\) suffices.
Here are some scenarios where approximation methods are useful:
- Calculating measurements in construction or design
- Estimating growth rates in population studies
- Predicting financial forecasts
- Performing engineering calculations
Being adept at knowing when and how to approximate is a vital skill when tackling exponential equations.