Discover answers to your most pressing questions at Westonci.ca, the ultimate Q&A platform that connects you with expert solutions. Experience the convenience of getting reliable answers to your questions from a vast network of knowledgeable experts. Get precise and detailed answers to your questions from a knowledgeable community of experts on our 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]
Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.