Answer :
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]
