Sure, let's solve the equation step-by-step:
Given equation:
[tex]\[ 16 \times 2^{(x+1)} = 4^x \times 8^{(1-x)} \][/tex]
First, we express all the constants (16, 4, and 8) with base 2:
- [tex]\(16 = 2^4\)[/tex]
- [tex]\(4 = 2^2\)[/tex]
- [tex]\(8 = 2^3\)[/tex]
Substitute these into the equation:
[tex]\[ 2^4 \times 2^{(x+1)} = (2^2)^x \times (2^3)^{(1-x)} \][/tex]
Simplify the exponents:
[tex]\[ 2^{4} \times 2^{x+1} = 2^{2x} \times 2^{3(1-x)} \][/tex]
Combine the exponents on each side:
[tex]\[ 2^{4+x+1} = 2^{2x + 3 - 3x} \][/tex]
Simplify the exponents again:
[tex]\[ 2^{5+x} = 2^{3 - x} \][/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ 5 + x = 3 - x \][/tex]
Solve for [tex]\(x\)[/tex]:
[tex]\[ 5 + x + x = 3 \][/tex]
[tex]\[ 5 + 2x = 3 \][/tex]
[tex]\[ 2x = 3 - 5 \][/tex]
[tex]\[ 2x = -2 \][/tex]
[tex]\[ x = -1 \][/tex]
So, one solution is:
[tex]\[ x = -1 \][/tex]
However, equations involving exponential functions can sometimes yield complex solutions. To find all solutions, we include the factor of [tex]\(2k\pi i\)[/tex] (where [tex]\(k\)[/tex] is an integer) to account for other potential solutions involving the complex logarithm.
Thus, we need to consider other possible solutions. To find the complete set analytically can be complicated, but the comprehensive solution should recognize complex results that arise from such equations. Here we consider the visible complex component directly without requiring re-evaluation:
Thus, the solutions are:
[tex]\[ x = -1 \][/tex]
[tex]\[ x = -1 + \frac{i \pi}{\ln(2)} \][/tex]
So, the final solutions are:
[tex]\[ x = -1 \quad \text{and} \quad x = -1 + \frac{i \pi}{\ln(2)} \][/tex]
These are the solutions to the given equation.
