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To solve the equation [tex]\( 25^x \cdot 5^{x+2} + 100 = 0 \)[/tex], we'll follow a structured approach to find the values of [tex]\( x \)[/tex]. Here are the steps to solve this equation:
1. Rewrite the given equation for clarity:
[tex]\[ 25^x \cdot 5^{x+2} + 100 = 0 \][/tex]
2. Simplify the exponents:
Recall that [tex]\( 25 = 5^2 \)[/tex]. Hence, we can rewrite [tex]\( 25^x \)[/tex] as [tex]\( (5^2)^x = 5^{2x} \)[/tex].
Substituting this into the equation gives us:
[tex]\[ 5^{2x} \cdot 5^{x+2} + 100 = 0 \][/tex]
3. Combine the exponents:
Using the property of exponents that states [tex]\( a^m \cdot a^n = a^{m+n} \)[/tex], we can combine the powers of 5:
[tex]\[ 5^{2x + (x+2)} + 100 = 0 \][/tex]
Simplifying the exponent, we get:
[tex]\[ 5^{3x+2} + 100 = 0 \][/tex]
4. Isolate the exponential term:
[tex]\[ 5^{3x+2} = -100 \][/tex]
5. Solve for [tex]\( x \)[/tex] using logarithms:
[tex]\[ 5^{3x+2} = -100 \][/tex]
Notice that it is not possible for any real number [tex]\( x \)[/tex] to satisfy this equation because the exponent of a positive base (like 5) to any real power is positive, and hence cannot equal the negative number [tex]\(-100\)[/tex].
However, considering complex solutions, [tex]\( 5^{3x+2} \)[/tex] must be equated to a complex logarithmic expression.
6. Equate the exponents:
To find the solutions in the complex plane, express [tex]\(-100\)[/tex] as a complex exponent. For a general base [tex]\( a \)[/tex] and exponent [tex]\( b \)[/tex]:
[tex]\[ a^b = e^{b \log(a)} \][/tex]
So, we rewrite:
[tex]\[ 5^{3x+2} = e^{(3x+2) \log(5)} \][/tex]
7. Express [tex]\(-100\)[/tex] in exponential form:
We can write [tex]\(-100\)[/tex] as follows:
[tex]\[ -100 = 100 \cdot e^{i \pi (2k + 1)} \quad \text{(since} \; e^{i\pi} = -1\text{ for integer k)} \][/tex]
8. Equate the exponents:
By matching the exponents, we get:
[tex]\[ 3x + 2 = \frac{\log(100) + i\pi (2k+1)}{\log(5)} \][/tex]
9. Solve for [tex]\( x \)[/tex]:
Isolate [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\log(100) + i\pi (2k+1)}{3 \log(5)} - \frac{2}{3} \][/tex]
Simplifying the logarithm:
[tex]\[ \log(100) = 2 \log(10) = 2 \log(10) = 2 \log(2 \cdot 5) = 2 (\log(2) + \log(5)) \][/tex]
Hence,
[tex]\[ x = \frac{2 (\log(2) + \log(5)) + i\pi (2k+1)}{3 \log(5)} - \frac{2}{3} \][/tex]
10. Simplify the expression:
Solving for [tex]\( x \)[/tex] yields the final complex solutions:
[tex]\[ x = \frac{2 \log(2) + i \pi(2k+1) + 2 \log(5)}{3 \log(5)} - \frac{2}{3} \][/tex]
This condenses to:
[tex]\[ x = \frac{2 \log(2) + 2 \log(5)}{3 \log(5)} + \frac{ i \pi (2k+1)}{3 \log(5)} = \frac{2 (\log(2) + \log(5))}{3 \log(5)} + \frac{ i \pi (2k+1)}{3 \log(5)} - \frac{2}{3} \][/tex]
Given full simplification for initial principal value by periodic addition we identify solutions to be:
[tex]\[ \frac{2 \log 2}{3 \log 5} + \frac{I \pi}{\log 5} \quad \text{or} \quad \frac{\log 4 - I \pi}{3 \log 5}, \quad \frac{\log 4 + I \pi}{3 \log 5} \][/tex]
using logs of number manipulations shows principal solutions variants as:
[tex]\((2(\log2)/3 + I\pi)/log5\)[/tex], [tex]\((\log 4 - I \pi)/(3 \log 5)\)[/tex], and, [tex]\(\log4 + I\pi/(3 \log 5)\)[/tex].
Thus, the solution for the equation is:
[tex]\[ x = \frac{(2 \log{2}) + i \pi}{3 \log{5}}, \frac{\log{4} + i \pi}{3 \log{5}}, \quad \frac{\log{4} - i \pi}{3 \log{5}} \][/tex]
1. Rewrite the given equation for clarity:
[tex]\[ 25^x \cdot 5^{x+2} + 100 = 0 \][/tex]
2. Simplify the exponents:
Recall that [tex]\( 25 = 5^2 \)[/tex]. Hence, we can rewrite [tex]\( 25^x \)[/tex] as [tex]\( (5^2)^x = 5^{2x} \)[/tex].
Substituting this into the equation gives us:
[tex]\[ 5^{2x} \cdot 5^{x+2} + 100 = 0 \][/tex]
3. Combine the exponents:
Using the property of exponents that states [tex]\( a^m \cdot a^n = a^{m+n} \)[/tex], we can combine the powers of 5:
[tex]\[ 5^{2x + (x+2)} + 100 = 0 \][/tex]
Simplifying the exponent, we get:
[tex]\[ 5^{3x+2} + 100 = 0 \][/tex]
4. Isolate the exponential term:
[tex]\[ 5^{3x+2} = -100 \][/tex]
5. Solve for [tex]\( x \)[/tex] using logarithms:
[tex]\[ 5^{3x+2} = -100 \][/tex]
Notice that it is not possible for any real number [tex]\( x \)[/tex] to satisfy this equation because the exponent of a positive base (like 5) to any real power is positive, and hence cannot equal the negative number [tex]\(-100\)[/tex].
However, considering complex solutions, [tex]\( 5^{3x+2} \)[/tex] must be equated to a complex logarithmic expression.
6. Equate the exponents:
To find the solutions in the complex plane, express [tex]\(-100\)[/tex] as a complex exponent. For a general base [tex]\( a \)[/tex] and exponent [tex]\( b \)[/tex]:
[tex]\[ a^b = e^{b \log(a)} \][/tex]
So, we rewrite:
[tex]\[ 5^{3x+2} = e^{(3x+2) \log(5)} \][/tex]
7. Express [tex]\(-100\)[/tex] in exponential form:
We can write [tex]\(-100\)[/tex] as follows:
[tex]\[ -100 = 100 \cdot e^{i \pi (2k + 1)} \quad \text{(since} \; e^{i\pi} = -1\text{ for integer k)} \][/tex]
8. Equate the exponents:
By matching the exponents, we get:
[tex]\[ 3x + 2 = \frac{\log(100) + i\pi (2k+1)}{\log(5)} \][/tex]
9. Solve for [tex]\( x \)[/tex]:
Isolate [tex]\( x \)[/tex]:
[tex]\[ x = \frac{\log(100) + i\pi (2k+1)}{3 \log(5)} - \frac{2}{3} \][/tex]
Simplifying the logarithm:
[tex]\[ \log(100) = 2 \log(10) = 2 \log(10) = 2 \log(2 \cdot 5) = 2 (\log(2) + \log(5)) \][/tex]
Hence,
[tex]\[ x = \frac{2 (\log(2) + \log(5)) + i\pi (2k+1)}{3 \log(5)} - \frac{2}{3} \][/tex]
10. Simplify the expression:
Solving for [tex]\( x \)[/tex] yields the final complex solutions:
[tex]\[ x = \frac{2 \log(2) + i \pi(2k+1) + 2 \log(5)}{3 \log(5)} - \frac{2}{3} \][/tex]
This condenses to:
[tex]\[ x = \frac{2 \log(2) + 2 \log(5)}{3 \log(5)} + \frac{ i \pi (2k+1)}{3 \log(5)} = \frac{2 (\log(2) + \log(5))}{3 \log(5)} + \frac{ i \pi (2k+1)}{3 \log(5)} - \frac{2}{3} \][/tex]
Given full simplification for initial principal value by periodic addition we identify solutions to be:
[tex]\[ \frac{2 \log 2}{3 \log 5} + \frac{I \pi}{\log 5} \quad \text{or} \quad \frac{\log 4 - I \pi}{3 \log 5}, \quad \frac{\log 4 + I \pi}{3 \log 5} \][/tex]
using logs of number manipulations shows principal solutions variants as:
[tex]\((2(\log2)/3 + I\pi)/log5\)[/tex], [tex]\((\log 4 - I \pi)/(3 \log 5)\)[/tex], and, [tex]\(\log4 + I\pi/(3 \log 5)\)[/tex].
Thus, the solution for the equation is:
[tex]\[ x = \frac{(2 \log{2}) + i \pi}{3 \log{5}}, \frac{\log{4} + i \pi}{3 \log{5}}, \quad \frac{\log{4} - i \pi}{3 \log{5}} \][/tex]
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