To solve the given equation [tex]\(4^x \times 16^{(x-1)} = 1\)[/tex], let's proceed step by step:
1. Rewrite the bases in terms of base 2:
- [tex]\( 4 \)[/tex] can be expressed as [tex]\(2^2\)[/tex], so [tex]\(4^x = (2^2)^x = 2^{2x}\)[/tex].
- [tex]\( 16 \)[/tex] can be expressed as [tex]\(2^4\)[/tex], so [tex]\(16^{(x-1)} = (2^4)^{(x-1)} = 2^{4(x-1)} = 2^{4x - 4}\)[/tex].
2. Substitute these into the equation:
[tex]\[
4^x \times 16^{(x-1)} = 2^{2x} \times 2^{4x-4} = 1.
\][/tex]
3. Combine the exponents with the same base:
[tex]\[
2^{2x} \times 2^{4x-4} = 2^{(2x + 4x - 4)} = 2^{6x-4}.
\][/tex]
4. Since the product of the exponents results in [tex]\(2^{6x-4}\)[/tex], we set this equal to 1:
- Recall that [tex]\(2^0 = 1\)[/tex].
[tex]\[
2^{6x-4} = 2^0.
\][/tex]
5. Equate the exponents:
[tex]\[
6x - 4 = 0.
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
[tex]\[
6x = 4 \implies x = \frac{4}{6} \implies x = \frac{2}{3}.
\][/tex]
Next, we also need to consider the presence of any complex solutions due to the periodic nature of the exponential function. Specifically, for any complex solution [tex]\(z\)[/tex]:
Given our original equation [tex]\(4^x \times 16^{(x-1)} - 1 = 0\)[/tex], we recognize that the equation can be expressed in complex form using logarithms and properties of exponentials. The general complex solutions involve logarithmic identities considering complex [tex]\(i\)[/tex] components and periodicity of the logarithm in the complex plane.
Thus, the set of solutions is:
[tex]\[
\left\{\frac{2}{3}, \frac{\log(4) - i\pi}{3 \log(2)}, \frac{\log(4) + i\pi}{3 \log(2)}, \frac{2}{3} - \frac{2 i \pi}{3 \log(2)}, \frac{2}{3} + \frac{2 i \pi}{3 \log(2)}, \frac{2}{3} + \frac{i \pi}{\log(2)}\right\}.
\][/tex]
Therefore, the complete set of solutions for [tex]\(x\)[/tex] includes both the real solution and complex solutions.
