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Sagot :
To solve the equation [tex]\(5^{3 r} = 5^{-2 r}\)[/tex], let’s follow these steps:
1. Set the exponents equal to each other: Since the bases are the same (both are 5), we can set the exponents equal to each other:
[tex]\[ 3r = -2r \][/tex]
2. Solve for [tex]\(r\)[/tex]: Now, solve the equation [tex]\(3r = -2r\)[/tex].
[tex]\[ 3r + 2r = 0 \][/tex]
Simplifying the left-hand side:
[tex]\[ 5r = 0 \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r = 0 \][/tex]
So, one solution is [tex]\(r = 0\)[/tex].
3. Consider the periodic nature of exponential functions: Exponential functions can have complex solutions due to their periodicity, especially when the exponential equation involves the same base raised to different power arguments that can be equivalent up to integer multiples of [tex]\(2\pi i\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit. In general, for [tex]\( a^b = a^c \)[/tex] where [tex]\( a \)[/tex] is a positive real number, [tex]\( b \)[/tex] and [tex]\( c \)[/tex] can be related through [tex]\(\log(a)\)[/tex] and complex multiples of [tex]\(2\pi i \)[/tex]:
[tex]\[ 3r = -2r + 2k\pi i / \log(5) \][/tex]
for integer [tex]\(k\)[/tex].
4. Solve the general complex solution:
[tex]\[ 3r + 2r = 2k\pi i / \log(5) \][/tex]
[tex]\[ 5r = 2k\pi i / \log(5) \][/tex]
[tex]\[ r = \frac{2k\pi i}{5 \log(5)} \][/tex]
This provides a family of solutions based on different values of [tex]\(k\)[/tex].
5. Identify specific solutions by substituting [tex]\(k = 0, \pm1, \pm2, \ldots\)[/tex]:
- For [tex]\( k = 0\)[/tex]:
[tex]\[ r = 0 \][/tex]
This is the real solution we already found.
- For [tex]\( k = 1\)[/tex]:
[tex]\[ r = \frac{2\pi i}{5 \log(5)} \][/tex]
- For [tex]\(k = -1\)[/tex]:
[tex]\[ r = -\frac{2\pi i}{5 \log(5)} \][/tex]
- For [tex]\( k = 2\)[/tex]:
[tex]\[ r = \frac{4\pi i}{5 \log(5)} \][/tex]
- For [tex]\( k = -2\)[/tex]:
[tex]\[ r = -\frac{4\pi i}{5 \log(5)} \][/tex]
Therefore, the complete set of solutions is:
[tex]\[ \begin{aligned} r &= 0, \\ r &= \frac{2\pi i}{5 \log(5)}, \\ r &= -\frac{2\pi i}{5 \log(5)}, \\ r &= \frac{4\pi i}{5 \log(5)}, \\ r &= -\frac{4\pi i}{5 \log(5)}. \end{aligned} \][/tex]
These are the detailed step-by-step solutions for the equation [tex]\(5^{3 r} = 5^{-2 r}\)[/tex].
1. Set the exponents equal to each other: Since the bases are the same (both are 5), we can set the exponents equal to each other:
[tex]\[ 3r = -2r \][/tex]
2. Solve for [tex]\(r\)[/tex]: Now, solve the equation [tex]\(3r = -2r\)[/tex].
[tex]\[ 3r + 2r = 0 \][/tex]
Simplifying the left-hand side:
[tex]\[ 5r = 0 \][/tex]
Solving for [tex]\(r\)[/tex]:
[tex]\[ r = 0 \][/tex]
So, one solution is [tex]\(r = 0\)[/tex].
3. Consider the periodic nature of exponential functions: Exponential functions can have complex solutions due to their periodicity, especially when the exponential equation involves the same base raised to different power arguments that can be equivalent up to integer multiples of [tex]\(2\pi i\)[/tex], where [tex]\(i\)[/tex] is the imaginary unit. In general, for [tex]\( a^b = a^c \)[/tex] where [tex]\( a \)[/tex] is a positive real number, [tex]\( b \)[/tex] and [tex]\( c \)[/tex] can be related through [tex]\(\log(a)\)[/tex] and complex multiples of [tex]\(2\pi i \)[/tex]:
[tex]\[ 3r = -2r + 2k\pi i / \log(5) \][/tex]
for integer [tex]\(k\)[/tex].
4. Solve the general complex solution:
[tex]\[ 3r + 2r = 2k\pi i / \log(5) \][/tex]
[tex]\[ 5r = 2k\pi i / \log(5) \][/tex]
[tex]\[ r = \frac{2k\pi i}{5 \log(5)} \][/tex]
This provides a family of solutions based on different values of [tex]\(k\)[/tex].
5. Identify specific solutions by substituting [tex]\(k = 0, \pm1, \pm2, \ldots\)[/tex]:
- For [tex]\( k = 0\)[/tex]:
[tex]\[ r = 0 \][/tex]
This is the real solution we already found.
- For [tex]\( k = 1\)[/tex]:
[tex]\[ r = \frac{2\pi i}{5 \log(5)} \][/tex]
- For [tex]\(k = -1\)[/tex]:
[tex]\[ r = -\frac{2\pi i}{5 \log(5)} \][/tex]
- For [tex]\( k = 2\)[/tex]:
[tex]\[ r = \frac{4\pi i}{5 \log(5)} \][/tex]
- For [tex]\( k = -2\)[/tex]:
[tex]\[ r = -\frac{4\pi i}{5 \log(5)} \][/tex]
Therefore, the complete set of solutions is:
[tex]\[ \begin{aligned} r &= 0, \\ r &= \frac{2\pi i}{5 \log(5)}, \\ r &= -\frac{2\pi i}{5 \log(5)}, \\ r &= \frac{4\pi i}{5 \log(5)}, \\ r &= -\frac{4\pi i}{5 \log(5)}. \end{aligned} \][/tex]
These are the detailed step-by-step solutions for the equation [tex]\(5^{3 r} = 5^{-2 r}\)[/tex].
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