Welcome to Westonci.ca, the place where your questions find answers from a community of knowledgeable experts. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
Of course! Let's solve the equation [tex]\( 128 = 4^{2x} \times 2^x \)[/tex] step-by-step.
1. Rewrite the bases in terms of powers of 2:
- We know that [tex]\( 4 = 2^2 \)[/tex]. Therefore, [tex]\( 4^{2x} = (2^2)^{2x} = 2^{4x} \)[/tex].
- Thus, the equation becomes:
[tex]\[ 128 = 2^{4x} \times 2^x \][/tex]
2. Combine the exponents on the right-hand side:
- Using the property of exponents [tex]\( a^m \times a^n = a^{m+n} \)[/tex], we can combine the exponents:
[tex]\[ 128 = 2^{4x + x} = 2^{5x} \][/tex]
3. Express 128 as a power of 2:
- We know that [tex]\( 128 = 2^7 \)[/tex]. So, the equation becomes:
[tex]\[ 2^7 = 2^{5x} \][/tex]
4. Set the exponents equal to each other:
- Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 7 = 5x \][/tex]
5. Solve for [tex]\( x \)[/tex]:
- Divide both sides of the equation by 5:
[tex]\[ x = \frac{7}{5} \][/tex]
So, one of the solutions for [tex]\( x \)[/tex] is [tex]\( x = \frac{7}{5} \)[/tex].
6. Consider the general solution for exponents involving complex numbers:
- To encompass all possible solutions, including complex numbers, we must remember that:
[tex]\[ 2^{5x} = 128 \][/tex]
- Taking the natural logarithm of both sides:
[tex]\[ \log(2^{5x}) = \log(128) \][/tex]
- Using the property of logarithms [tex]\( \log(a^b) = b\log(a) \)[/tex]:
[tex]\[ 5x \log(2) = \log(128) \][/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\log(128)}{5 \log(2)} \][/tex]
- However, for a complete solution, we consider the periodicity of the logarithm in the complex plane:
[tex]\[ x = \frac{\log(128) + 2k\pi i}{5 \log(2)} \][/tex]
- Simplifying the constant [tex]\(\log(128)\)[/tex]:
[tex]\[ \log(128) = 7 \log(2) \][/tex]
Thus, the general form is:
[tex]\[ x = \frac{7 \log(2) + 2k\pi i}{5\log(2)} = \frac{7}{5} + \frac{2k\pi i}{5\log(2)} \][/tex]
Summarizing the solutions:
- Real part: [tex]\( x = \frac{7}{5} \)[/tex]
- Complex parts: [tex]\( x = \frac{7}{5} + \frac{2k\pi i}{5\log(2)} \text{ for integer } k \)[/tex]
These represent all possible values of [tex]\( x \)[/tex] that satisfy the original equation. The exact solutions are:
[tex]\[ x = \frac{7}{5}, \quad x = \frac{\log(128) - 4i\pi}{5\log(2)}, \quad x = \frac{\log(128) - 2i\pi}{5\log(2)}, \quad x = \frac{\log(128) + 2i\pi}{5\log(2)}, \quad x = \frac{\log(128) + 4i\pi}{5\log(2)} \][/tex]
1. Rewrite the bases in terms of powers of 2:
- We know that [tex]\( 4 = 2^2 \)[/tex]. Therefore, [tex]\( 4^{2x} = (2^2)^{2x} = 2^{4x} \)[/tex].
- Thus, the equation becomes:
[tex]\[ 128 = 2^{4x} \times 2^x \][/tex]
2. Combine the exponents on the right-hand side:
- Using the property of exponents [tex]\( a^m \times a^n = a^{m+n} \)[/tex], we can combine the exponents:
[tex]\[ 128 = 2^{4x + x} = 2^{5x} \][/tex]
3. Express 128 as a power of 2:
- We know that [tex]\( 128 = 2^7 \)[/tex]. So, the equation becomes:
[tex]\[ 2^7 = 2^{5x} \][/tex]
4. Set the exponents equal to each other:
- Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 7 = 5x \][/tex]
5. Solve for [tex]\( x \)[/tex]:
- Divide both sides of the equation by 5:
[tex]\[ x = \frac{7}{5} \][/tex]
So, one of the solutions for [tex]\( x \)[/tex] is [tex]\( x = \frac{7}{5} \)[/tex].
6. Consider the general solution for exponents involving complex numbers:
- To encompass all possible solutions, including complex numbers, we must remember that:
[tex]\[ 2^{5x} = 128 \][/tex]
- Taking the natural logarithm of both sides:
[tex]\[ \log(2^{5x}) = \log(128) \][/tex]
- Using the property of logarithms [tex]\( \log(a^b) = b\log(a) \)[/tex]:
[tex]\[ 5x \log(2) = \log(128) \][/tex]
- Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\log(128)}{5 \log(2)} \][/tex]
- However, for a complete solution, we consider the periodicity of the logarithm in the complex plane:
[tex]\[ x = \frac{\log(128) + 2k\pi i}{5 \log(2)} \][/tex]
- Simplifying the constant [tex]\(\log(128)\)[/tex]:
[tex]\[ \log(128) = 7 \log(2) \][/tex]
Thus, the general form is:
[tex]\[ x = \frac{7 \log(2) + 2k\pi i}{5\log(2)} = \frac{7}{5} + \frac{2k\pi i}{5\log(2)} \][/tex]
Summarizing the solutions:
- Real part: [tex]\( x = \frac{7}{5} \)[/tex]
- Complex parts: [tex]\( x = \frac{7}{5} + \frac{2k\pi i}{5\log(2)} \text{ for integer } k \)[/tex]
These represent all possible values of [tex]\( x \)[/tex] that satisfy the original equation. The exact solutions are:
[tex]\[ x = \frac{7}{5}, \quad x = \frac{\log(128) - 4i\pi}{5\log(2)}, \quad x = \frac{\log(128) - 2i\pi}{5\log(2)}, \quad x = \frac{\log(128) + 2i\pi}{5\log(2)}, \quad x = \frac{\log(128) + 4i\pi}{5\log(2)} \][/tex]
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.
