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To solve for [tex]\( x \)[/tex] in the equation
[tex]\[ 32^{(x-3)} \div 8^{(x-4)} = 64 \div 2^x, \][/tex]
we start by simplifying both sides of the equation.
1. Express all numbers in terms of base 2:
- [tex]\( 32 = 2^5 \)[/tex]
- [tex]\( 8 = 2^3 \)[/tex]
- [tex]\( 64 = 2^6 \)[/tex]
2. Substitute these bases into the equation:
[tex]\[ (2^5)^{(x-3)} \div (2^3)^{(x-4)} = \frac{2^6}{2^x} \][/tex]
3. Simplify the exponents on the left side:
[tex]\[ 2^{5(x-3)} \div 2^{3(x-4)} = 2^{5x-15} \div 2^{3x-12} \][/tex]
4. Apply the quotient rule for exponents, which states [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex]:
[tex]\[ \frac{2^{5x-15}}{2^{3x-12}} = 2^{(5x-15) - (3x-12)} \][/tex]
5. Simplify the exponent:
[tex]\[ 2^{5x-15-3x+12} = 2^{2x - 3} \][/tex]
So, we now have:
[tex]\[ 2^{2x - 3} = \frac{2^6}{2^x} \][/tex]
6. Simplify the right side using the quotient rule:
[tex]\[ 2^{2x - 3} = 2^{6 - x} \][/tex]
7. Since the bases are the same, we can equate the exponents:
[tex]\[ 2x - 3 = 6 - x \][/tex]
8. Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 3 + x = 6 \][/tex]
[tex]\[ 3x - 3 = 6 \][/tex]
[tex]\[ 3x = 9 \][/tex]
[tex]\[ x = 3 \][/tex]
Therefore, one solution for [tex]\( x \)[/tex] is [tex]\( x = 3 \)[/tex].
Next, let's discuss the complex solutions. The equation can also be solved using the properties of logarithms for complex numbers, which involves considering the periodicity of the logarithm function.
From our algebraic work using logs:
Given our simplified equation:
[tex]\[ 2^{2x - 3} = 2^{6 - x} \][/tex]
Taking the natural logarithm on both sides yields:
[tex]\[ \ln(2^{2x - 3}) = \ln(2^{6 - x}) \][/tex]
Using the property [tex]\(\ln(a^b) = b\ln(a)\)[/tex]:
[tex]\[ (2x - 3)\ln(2) = (6 - x)\ln(2) \][/tex]
Again, equate the exponents:
[tex]\[ 2x - 3 - 6 + x = 2n\pi i \ln(2) / \ln(2) = 2n\pi i \ln(2 /\ln(2)) 2x - 3 = (6 - x)+2\pi i So, solving general log equation, we get complex solutions: \][/tex]
The solutions include:
[tex]\[x = 3, \frac{\log(512) \pm 2i\pi}{3 \log(2)} ] Therefore, the solutions to the equation \( 32^{(x-3)} \div 8^{(x-4)} = 64 \div 2^x \) are: \[ x = 3,\quad (5/9, 2\imath\pm \pi ) which can simplify to\][/tex]
\[ x = 3 (log(2)/3) as desired)."
[tex]\[ 32^{(x-3)} \div 8^{(x-4)} = 64 \div 2^x, \][/tex]
we start by simplifying both sides of the equation.
1. Express all numbers in terms of base 2:
- [tex]\( 32 = 2^5 \)[/tex]
- [tex]\( 8 = 2^3 \)[/tex]
- [tex]\( 64 = 2^6 \)[/tex]
2. Substitute these bases into the equation:
[tex]\[ (2^5)^{(x-3)} \div (2^3)^{(x-4)} = \frac{2^6}{2^x} \][/tex]
3. Simplify the exponents on the left side:
[tex]\[ 2^{5(x-3)} \div 2^{3(x-4)} = 2^{5x-15} \div 2^{3x-12} \][/tex]
4. Apply the quotient rule for exponents, which states [tex]\(\frac{a^m}{a^n} = a^{m-n}\)[/tex]:
[tex]\[ \frac{2^{5x-15}}{2^{3x-12}} = 2^{(5x-15) - (3x-12)} \][/tex]
5. Simplify the exponent:
[tex]\[ 2^{5x-15-3x+12} = 2^{2x - 3} \][/tex]
So, we now have:
[tex]\[ 2^{2x - 3} = \frac{2^6}{2^x} \][/tex]
6. Simplify the right side using the quotient rule:
[tex]\[ 2^{2x - 3} = 2^{6 - x} \][/tex]
7. Since the bases are the same, we can equate the exponents:
[tex]\[ 2x - 3 = 6 - x \][/tex]
8. Solve for [tex]\( x \)[/tex]:
[tex]\[ 2x - 3 + x = 6 \][/tex]
[tex]\[ 3x - 3 = 6 \][/tex]
[tex]\[ 3x = 9 \][/tex]
[tex]\[ x = 3 \][/tex]
Therefore, one solution for [tex]\( x \)[/tex] is [tex]\( x = 3 \)[/tex].
Next, let's discuss the complex solutions. The equation can also be solved using the properties of logarithms for complex numbers, which involves considering the periodicity of the logarithm function.
From our algebraic work using logs:
Given our simplified equation:
[tex]\[ 2^{2x - 3} = 2^{6 - x} \][/tex]
Taking the natural logarithm on both sides yields:
[tex]\[ \ln(2^{2x - 3}) = \ln(2^{6 - x}) \][/tex]
Using the property [tex]\(\ln(a^b) = b\ln(a)\)[/tex]:
[tex]\[ (2x - 3)\ln(2) = (6 - x)\ln(2) \][/tex]
Again, equate the exponents:
[tex]\[ 2x - 3 - 6 + x = 2n\pi i \ln(2) / \ln(2) = 2n\pi i \ln(2 /\ln(2)) 2x - 3 = (6 - x)+2\pi i So, solving general log equation, we get complex solutions: \][/tex]
The solutions include:
[tex]\[x = 3, \frac{\log(512) \pm 2i\pi}{3 \log(2)} ] Therefore, the solutions to the equation \( 32^{(x-3)} \div 8^{(x-4)} = 64 \div 2^x \) are: \[ x = 3,\quad (5/9, 2\imath\pm \pi ) which can simplify to\][/tex]
\[ x = 3 (log(2)/3) as desired)."
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