Answer :
To solve the equation [tex]\(16^x - 5 \times 4^{x+1} + 64 = 0\)[/tex], let's closely examine it step by step.
1. Rewrite the equation using powers of 2:
Observe that [tex]\(16\)[/tex] and [tex]\(4\)[/tex] can be expressed as powers of 2:
[tex]\[ 16 = 2^4 \quad \text{and} \quad 4 = 2^2 \][/tex]
Hence, we can rewrite the equation as:
[tex]\[ (2^4)^x - 5 \times (2^2)^{x+1} + 64 = 0 \][/tex]
2. Simplify the exponents:
Simplify the exponents using the power rules [tex]\( (a^m)^n = a^{mn} \)[/tex]:
[tex]\[ 2^{4x} - 5 \times 2^{2(x+1)} + 64 = 0 \][/tex]
Next, distribute the exponent in the second term:
[tex]\[ 2^{4x} - 5 \times 2^{2x+2} + 64 = 0 \][/tex]
3. Introduce a substitution:
To create a quadratic form, let [tex]\(y = 2^{2x}\)[/tex]. Notice that [tex]\(2^{4x} = (2^{2x})^2 = y^2\)[/tex] and [tex]\(2^{2x+2} = 2^2 \times 2^{2x} = 4y\)[/tex]. Substitute these into the equation:
[tex]\[ y^2 - 5 \times 4y + 64 = 0 \][/tex]
Simplify:
[tex]\[ y^2 - 20y + 64 = 0 \][/tex]
4. Solve the quadratic equation:
Solve the quadratic equation [tex]\(y^2 - 20y + 64 = 0\)[/tex] using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -20\)[/tex], and [tex]\(c = 64\)[/tex]:
[tex]\[ y = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{20 \pm \sqrt{400 - 256}}{2} \][/tex]
[tex]\[ y = \frac{20 \pm \sqrt{144}}{2} \][/tex]
[tex]\[ y = \frac{20 \pm 12}{2} \][/tex]
This gives us two solutions:
[tex]\[ y = \frac{32}{2} = 16 \quad \text{and} \quad y = \frac{8}{2} = 4 \][/tex]
5. Convert back to [tex]\(x\)[/tex]:
Recall [tex]\(y = 2^{2x}\)[/tex], so we have two cases:
[tex]\[ 2^{2x} = 16 \quad \text{and} \quad 2^{2x} = 4 \][/tex]
Write these in exponential form:
[tex]\[ 2^{2x} = 2^4 \quad \text{and} \quad 2^{2x} = 2^2 \][/tex]
Equate the exponents:
[tex]\[ 2x = 4 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ 2x = 2 \quad \Rightarrow \quad x = 1 \][/tex]
6. Consider complex solutions:
The equations [tex]\(2^{2x} = 16\)[/tex] and [tex]\(2^{2x} = 4\)[/tex] could also have complex logarithmic solutions. Solving these using properties of logarithms and considering the periodicity of complex logarithms, we can determine the additional complex solutions:
[tex]\[ x = \frac{\log_e(4) + i\pi}{\log_e(2)}, \][/tex]
and
[tex]\[ x = 1 + \frac{i\pi}{\log_e(2)}. \][/tex]
Therefore, the complete set of solutions for the equation [tex]\(16^x - 5 \times 4^{x+1} + 64 = 0\)[/tex] is:
[tex]\[ x = 1, \, x = 2, \, x = \frac{\log(4) + I\pi}{\log(2)}, \, \text{and} \, x = 1 + \frac{I\pi}{\log(2)}. \][/tex]
1. Rewrite the equation using powers of 2:
Observe that [tex]\(16\)[/tex] and [tex]\(4\)[/tex] can be expressed as powers of 2:
[tex]\[ 16 = 2^4 \quad \text{and} \quad 4 = 2^2 \][/tex]
Hence, we can rewrite the equation as:
[tex]\[ (2^4)^x - 5 \times (2^2)^{x+1} + 64 = 0 \][/tex]
2. Simplify the exponents:
Simplify the exponents using the power rules [tex]\( (a^m)^n = a^{mn} \)[/tex]:
[tex]\[ 2^{4x} - 5 \times 2^{2(x+1)} + 64 = 0 \][/tex]
Next, distribute the exponent in the second term:
[tex]\[ 2^{4x} - 5 \times 2^{2x+2} + 64 = 0 \][/tex]
3. Introduce a substitution:
To create a quadratic form, let [tex]\(y = 2^{2x}\)[/tex]. Notice that [tex]\(2^{4x} = (2^{2x})^2 = y^2\)[/tex] and [tex]\(2^{2x+2} = 2^2 \times 2^{2x} = 4y\)[/tex]. Substitute these into the equation:
[tex]\[ y^2 - 5 \times 4y + 64 = 0 \][/tex]
Simplify:
[tex]\[ y^2 - 20y + 64 = 0 \][/tex]
4. Solve the quadratic equation:
Solve the quadratic equation [tex]\(y^2 - 20y + 64 = 0\)[/tex] using the quadratic formula [tex]\(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = -20\)[/tex], and [tex]\(c = 64\)[/tex]:
[tex]\[ y = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 1 \cdot 64}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{20 \pm \sqrt{400 - 256}}{2} \][/tex]
[tex]\[ y = \frac{20 \pm \sqrt{144}}{2} \][/tex]
[tex]\[ y = \frac{20 \pm 12}{2} \][/tex]
This gives us two solutions:
[tex]\[ y = \frac{32}{2} = 16 \quad \text{and} \quad y = \frac{8}{2} = 4 \][/tex]
5. Convert back to [tex]\(x\)[/tex]:
Recall [tex]\(y = 2^{2x}\)[/tex], so we have two cases:
[tex]\[ 2^{2x} = 16 \quad \text{and} \quad 2^{2x} = 4 \][/tex]
Write these in exponential form:
[tex]\[ 2^{2x} = 2^4 \quad \text{and} \quad 2^{2x} = 2^2 \][/tex]
Equate the exponents:
[tex]\[ 2x = 4 \quad \Rightarrow \quad x = 2 \][/tex]
[tex]\[ 2x = 2 \quad \Rightarrow \quad x = 1 \][/tex]
6. Consider complex solutions:
The equations [tex]\(2^{2x} = 16\)[/tex] and [tex]\(2^{2x} = 4\)[/tex] could also have complex logarithmic solutions. Solving these using properties of logarithms and considering the periodicity of complex logarithms, we can determine the additional complex solutions:
[tex]\[ x = \frac{\log_e(4) + i\pi}{\log_e(2)}, \][/tex]
and
[tex]\[ x = 1 + \frac{i\pi}{\log_e(2)}. \][/tex]
Therefore, the complete set of solutions for the equation [tex]\(16^x - 5 \times 4^{x+1} + 64 = 0\)[/tex] is:
[tex]\[ x = 1, \, x = 2, \, x = \frac{\log(4) + I\pi}{\log(2)}, \, \text{and} \, x = 1 + \frac{I\pi}{\log(2)}. \][/tex]
