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Sagot :
Sure! Let's match each equation with its solution.
1. Given the equation [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex], the solution is [tex]\( x = 3 \)[/tex].
2. Given the equation [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex], the solutions are [tex]\( x = 1.2 \)[/tex] and [tex]\( x = -1.0 \)[/tex].
3. Given the equation [tex]\(e^{x^2} = e^{4x+5}\)[/tex], the solutions are [tex]\( x = 5.0 \)[/tex] and [tex]\( x = -1.0 \)[/tex].
4. Given the equation [tex]\(\log(x-1) + \log(5x) = 2\)[/tex], the solutions are [tex]\( x = 5.0 \)[/tex] and [tex]\( x = -4.0 \)[/tex].
Here are the matches:
- [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex] only [tex]\(\rightarrow\)[/tex] only 3
- [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 1.2)\)[/tex]
- [tex]\(e^{x^2} = e^{4x+5}\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 5.0)\)[/tex]
- [tex]\(\log(x-1) + \log(5x) = 2\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\(5\)[/tex]
Hence:
- [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex] [tex]\(\rightarrow\)[/tex] only 3
- [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 1.2)\)[/tex]
- [tex]\(e^{x^2} = e^{4x+5}\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 5.0)\)[/tex]
- [tex]\(\log(x-1) + \log(5x) = 2\)[/tex] [tex]\(\rightarrow\)[/tex] 5
1. Given the equation [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex], the solution is [tex]\( x = 3 \)[/tex].
2. Given the equation [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex], the solutions are [tex]\( x = 1.2 \)[/tex] and [tex]\( x = -1.0 \)[/tex].
3. Given the equation [tex]\(e^{x^2} = e^{4x+5}\)[/tex], the solutions are [tex]\( x = 5.0 \)[/tex] and [tex]\( x = -1.0 \)[/tex].
4. Given the equation [tex]\(\log(x-1) + \log(5x) = 2\)[/tex], the solutions are [tex]\( x = 5.0 \)[/tex] and [tex]\( x = -4.0 \)[/tex].
Here are the matches:
- [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex] only [tex]\(\rightarrow\)[/tex] only 3
- [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 1.2)\)[/tex]
- [tex]\(e^{x^2} = e^{4x+5}\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 5.0)\)[/tex]
- [tex]\(\log(x-1) + \log(5x) = 2\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\(5\)[/tex]
Hence:
- [tex]\(\ln(x+5) = \ln(x-1) + \ln(x+1)\)[/tex] [tex]\(\rightarrow\)[/tex] only 3
- [tex]\(\log_4(5x^2+2) = \log_4(x+8)\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 1.2)\)[/tex]
- [tex]\(e^{x^2} = e^{4x+5}\)[/tex] [tex]\(\rightarrow\)[/tex] [tex]\((-1, 5.0)\)[/tex]
- [tex]\(\log(x-1) + \log(5x) = 2\)[/tex] [tex]\(\rightarrow\)[/tex] 5
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