Answer :
Let's solve each equation one by one step-by-step:
### Equation 1: [tex]\(\log (x-1) + \log (5x) = 2\)[/tex]
Using the properties of logarithms, we combine the logs:
[tex]\[ \log((x-1) \cdot 5x) = 2 \][/tex]
[tex]\[ \log(5x^2 - 5x) = 2 \][/tex]
Converting from logarithmic form to exponential form:
[tex]\[ 5x^2 - 5x = 10^2 \][/tex]
[tex]\[ 5x^2 - 5x = 100 \][/tex]
Solving the quadratic equation:
[tex]\[ 5x^2 - 5x - 100 = 0 \][/tex]
[tex]\[ x^2 - x - 20 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 1, \; b = -1, \; c = -20 \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 80}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 9}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{10}{2} = 5 \][/tex]
[tex]\[ x = \frac{-8}{2} = -4 \][/tex]
Since [tex]\(x\)[/tex] must satisfy the domain of the logarithms (i.e., [tex]\(x-1 > 0\)[/tex] and [tex]\(5x > 0\)[/tex]), [tex]\(x = -4\)[/tex] is not valid. Therefore, the solution is:
[tex]\[ x = 5 \][/tex]
### Equation 2: [tex]\(\log_4(5x^2 + 2) = \log_4(x + 8)\)[/tex]
Since the logarithms have the same base, we can set the arguments equal to each other:
[tex]\[ 5x^2 + 2 = x + 8 \][/tex]
Rearranging the equation:
[tex]\[ 5x^2 + 2 - x - 8 = 0 \][/tex]
[tex]\[ 5x^2 - x - 6 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 5, \; b = -1, \; c = -6 \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 120}}{10} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{121}}{10} \][/tex]
[tex]\[ x = \frac{1 \pm 11}{10} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{12}{10} = 1.2 \][/tex]
[tex]\[ x = \frac{-10}{10} = -1 \][/tex]
[tex]\(x\)[/tex] must be greater than [tex]\(\frac{-8}{5}\)[/tex] since [tex]\(5x^2 + 2 > 0 \geq x + 8\)[/tex].
Thus, [tex]\(x = -1 \)[/tex] will not satisfy this domain.
Therefore, the valid solution is:
[tex]\[ x = 1.2 \][/tex]
### Equation 3: [tex]\(\ln (x+5) = \ln (x-1) + \ln (x+1)\)[/tex]
Using the properties of logarithms, we combine the logs on the right-hand side:
[tex]\[ \ln (x+5) = \ln ((x-1)(x+1)) \][/tex]
[tex]\[ \ln (x+5) = \ln (x^2 - 1) \][/tex]
Equating the arguments of the natural logs:
[tex]\[ x + 5 = x^2 - 1 \][/tex]
Rearranging the equation:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 1, \; b = -1, \; c = -6 \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 24}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 5}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{6}{2} = 3 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
[tex]\(x\)[/tex] must be greater than 1 in the domain [tex]\(x-1 >0\)[/tex].
Thus, [tex]\(x = -2 \)[/tex] will not satisfy this domain.
Therefore the valid solution is:
[tex]\[ x = 3 \][/tex]
### Equation 4: [tex]\(e^{x^2} = e^{4x + 5}\)[/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ x^2 = 4x + 5 \][/tex]
Rearranging the equation:
[tex]\[ x^2 - 4x - 5 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 1, \; b = -4, \; c = -5 \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 + 20}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 6}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{10}{2} = 5 \][/tex]
[tex]\[ x = \frac{-2}{2} = -1 \][/tex]
Hence, the solutions are:
[tex]\[ x = 5 \; \text{and} \; x = -1 \][/tex]
### Matching Solutions:
[tex]\[ \begin{array}{lcl} \log (x-1)+\log 5x=2 & : & x = 5 \\ \log_4(5x^2 + 2) = \log_4(x + 8) & : & x = 1.2 \\ \ln (x+5) = \ln (x-1) + \ln (x+1) & : & x = 3 \\ e^{x^2} = e^{4x+5} & : & x = 5 \; \text{and} \; x = -1 \\ \end{array} \][/tex]
These matchings pair the equations with their respective solutions effectively.
### Equation 1: [tex]\(\log (x-1) + \log (5x) = 2\)[/tex]
Using the properties of logarithms, we combine the logs:
[tex]\[ \log((x-1) \cdot 5x) = 2 \][/tex]
[tex]\[ \log(5x^2 - 5x) = 2 \][/tex]
Converting from logarithmic form to exponential form:
[tex]\[ 5x^2 - 5x = 10^2 \][/tex]
[tex]\[ 5x^2 - 5x = 100 \][/tex]
Solving the quadratic equation:
[tex]\[ 5x^2 - 5x - 100 = 0 \][/tex]
[tex]\[ x^2 - x - 20 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 1, \; b = -1, \; c = -20 \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 80}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{81}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 9}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{10}{2} = 5 \][/tex]
[tex]\[ x = \frac{-8}{2} = -4 \][/tex]
Since [tex]\(x\)[/tex] must satisfy the domain of the logarithms (i.e., [tex]\(x-1 > 0\)[/tex] and [tex]\(5x > 0\)[/tex]), [tex]\(x = -4\)[/tex] is not valid. Therefore, the solution is:
[tex]\[ x = 5 \][/tex]
### Equation 2: [tex]\(\log_4(5x^2 + 2) = \log_4(x + 8)\)[/tex]
Since the logarithms have the same base, we can set the arguments equal to each other:
[tex]\[ 5x^2 + 2 = x + 8 \][/tex]
Rearranging the equation:
[tex]\[ 5x^2 + 2 - x - 8 = 0 \][/tex]
[tex]\[ 5x^2 - x - 6 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 5, \; b = -1, \; c = -6 \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 120}}{10} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{121}}{10} \][/tex]
[tex]\[ x = \frac{1 \pm 11}{10} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{12}{10} = 1.2 \][/tex]
[tex]\[ x = \frac{-10}{10} = -1 \][/tex]
[tex]\(x\)[/tex] must be greater than [tex]\(\frac{-8}{5}\)[/tex] since [tex]\(5x^2 + 2 > 0 \geq x + 8\)[/tex].
Thus, [tex]\(x = -1 \)[/tex] will not satisfy this domain.
Therefore, the valid solution is:
[tex]\[ x = 1.2 \][/tex]
### Equation 3: [tex]\(\ln (x+5) = \ln (x-1) + \ln (x+1)\)[/tex]
Using the properties of logarithms, we combine the logs on the right-hand side:
[tex]\[ \ln (x+5) = \ln ((x-1)(x+1)) \][/tex]
[tex]\[ \ln (x+5) = \ln (x^2 - 1) \][/tex]
Equating the arguments of the natural logs:
[tex]\[ x + 5 = x^2 - 1 \][/tex]
Rearranging the equation:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 1, \; b = -1, \; c = -6 \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 24}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 5}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{6}{2} = 3 \][/tex]
[tex]\[ x = \frac{-4}{2} = -2 \][/tex]
[tex]\(x\)[/tex] must be greater than 1 in the domain [tex]\(x-1 >0\)[/tex].
Thus, [tex]\(x = -2 \)[/tex] will not satisfy this domain.
Therefore the valid solution is:
[tex]\[ x = 3 \][/tex]
### Equation 4: [tex]\(e^{x^2} = e^{4x + 5}\)[/tex]
Since the bases are the same, we can set the exponents equal to each other:
[tex]\[ x^2 = 4x + 5 \][/tex]
Rearranging the equation:
[tex]\[ x^2 - 4x - 5 = 0 \][/tex]
Using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex]:
[tex]\[ a = 1, \; b = -4, \; c = -5 \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 + 20}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{36}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 6}{2} \][/tex]
So, the solutions are:
[tex]\[ x = \frac{10}{2} = 5 \][/tex]
[tex]\[ x = \frac{-2}{2} = -1 \][/tex]
Hence, the solutions are:
[tex]\[ x = 5 \; \text{and} \; x = -1 \][/tex]
### Matching Solutions:
[tex]\[ \begin{array}{lcl} \log (x-1)+\log 5x=2 & : & x = 5 \\ \log_4(5x^2 + 2) = \log_4(x + 8) & : & x = 1.2 \\ \ln (x+5) = \ln (x-1) + \ln (x+1) & : & x = 3 \\ e^{x^2} = e^{4x+5} & : & x = 5 \; \text{and} \; x = -1 \\ \end{array} \][/tex]
These matchings pair the equations with their respective solutions effectively.
