Answer :
Let's solve each equation step-by-step and find the correct solutions.
### Equation 1: [tex]\(\log (x-1) + \log 5x = 2\)[/tex]
Using the properties of logarithms:
[tex]\[ \log (x-1) + \log 5x = \log ((x-1) \cdot 5x) = 2 \][/tex]
[tex]\[ \log (5x(x-1)) = 2 \][/tex]
Convert the logarithmic equation to its exponential form:
[tex]\[ 5x(x-1) = 10^2 = 100 \][/tex]
[tex]\[ 5x^2 - 5x - 100 = 0 \][/tex]
Divide through by 5:
[tex]\[ x^2 - x - 20 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x - 5)(x + 4) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = 5 \text{ and } x = -4 \][/tex]
However, we must check for the domain (logarithms require positive arguments). For [tex]\(x = -4\)[/tex], [tex]\(\log (x-1)\)[/tex] and [tex]\(\log 5x\)[/tex] are undefined. Hence, the only valid solution here is:
[tex]\[ \boxed{\text{only 5}} \][/tex]
### Equation 2: [tex]\(\log_4(5x^2 + 2) = \log_4(x + 8)\)[/tex]
Since the bases and the logarithms are the same, we can equate their arguments directly:
[tex]\[ 5x^2 + 2 = x + 8 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ 5x^2 - x - 6 = 0 \][/tex]
Factor:
[tex]\[ (5x + 3)(x - 2) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = -\frac{3}{5} \text{ and } x = 2 \][/tex]
Check the domain: [tex]\(5x^2 + 2\)[/tex] and [tex]\(x + 8\)[/tex] must be positive.
For [tex]\(x = -\frac{3}{5}\)[/tex], both terms are positive:
[tex]\[ 5\left(-\frac{3}{5}\right)^2 + 2 > 0 \quad \text{and} \quad -\frac{3}{5} + 8 > 0 \][/tex]
For [tex]\(x = 2\)[/tex], both terms are positive:
[tex]\[ 5(2)^2 + 2 > 0 \quad \text{and} \quad 2 + 8 > 0 \][/tex]
Thus, both solutions are valid. So the completed answer is:
[tex]\[ \boxed{-\frac{3}{5} \text{ and } 2} \][/tex]
### Equation 3: [tex]\(e^{x^2} = e^{4x + 5}\)[/tex]
Since the bases are the same, we equate the exponents:
[tex]\[ x^2 = 4x + 5 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - 4x - 5 = 0 \][/tex]
Factor:
[tex]\[ (x - 5)(x + 1) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = 5 \text{ and } x = -1 \][/tex]
So the completed answer is:
[tex]\[ \boxed{-1 \text{ and } 5} \][/tex]
### Equation 4: [tex]\(\ln (x + 5) = \ln (x - 1) + \ln (x + 1)\)[/tex]
Use the properties of logarithms to combine the right-hand side:
[tex]\[ \ln (x + 5) = \ln ((x - 1)(x + 1)) \][/tex]
Thus:
[tex]\[ x + 5 = (x - 1)(x + 1) \][/tex]
Expand and simplify:
[tex]\[ x + 5 = x^2 - 1 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Factor:
[tex]\[ (x - \frac{6}{5})(x + \frac{1}{5}) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{6}{5} \text{ and } x = -\frac{1}{5} \][/tex]
Check for the domain: With natural logarithms, the arguments must be positive.
For [tex]\(x = \frac{6}{5}\)[/tex], [tex]\(x + 5 > 0, x - 1 > 0, x + 1 > 0\)[/tex].
For [tex]\(x = -\frac{1}{5}\)[/tex], [tex]\(x + 5 > 0\)[/tex], but [tex]\(x - 1 < 0\)[/tex], making it invalid.
Thus, the valid solution is:
[tex]\[ \boxed{\frac{6}{5}} \][/tex]
Matching each solution to its respective equation:
[tex]\[ \begin{array}{c} \log (x - 1) + \log 5x = 2 \quad \rightarrow \quad \boxed{\text{only 5}} \\ \log_4(5x^2 + 2) = \log_4(x + 8) \quad \rightarrow \quad \boxed{-\frac{3}{5} \text{ and } 2} \\ e^{x^2} = e^{4x + 5} \quad \rightarrow \quad \boxed{-1 \text{ and } 5} \\ \ln (x + 5) = \ln (x - 1) + \ln (x + 1) \quad \rightarrow \quad \boxed{\frac{6}{5}} \end{array} \][/tex]
### Equation 1: [tex]\(\log (x-1) + \log 5x = 2\)[/tex]
Using the properties of logarithms:
[tex]\[ \log (x-1) + \log 5x = \log ((x-1) \cdot 5x) = 2 \][/tex]
[tex]\[ \log (5x(x-1)) = 2 \][/tex]
Convert the logarithmic equation to its exponential form:
[tex]\[ 5x(x-1) = 10^2 = 100 \][/tex]
[tex]\[ 5x^2 - 5x - 100 = 0 \][/tex]
Divide through by 5:
[tex]\[ x^2 - x - 20 = 0 \][/tex]
Factor the quadratic equation:
[tex]\[ (x - 5)(x + 4) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = 5 \text{ and } x = -4 \][/tex]
However, we must check for the domain (logarithms require positive arguments). For [tex]\(x = -4\)[/tex], [tex]\(\log (x-1)\)[/tex] and [tex]\(\log 5x\)[/tex] are undefined. Hence, the only valid solution here is:
[tex]\[ \boxed{\text{only 5}} \][/tex]
### Equation 2: [tex]\(\log_4(5x^2 + 2) = \log_4(x + 8)\)[/tex]
Since the bases and the logarithms are the same, we can equate their arguments directly:
[tex]\[ 5x^2 + 2 = x + 8 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ 5x^2 - x - 6 = 0 \][/tex]
Factor:
[tex]\[ (5x + 3)(x - 2) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = -\frac{3}{5} \text{ and } x = 2 \][/tex]
Check the domain: [tex]\(5x^2 + 2\)[/tex] and [tex]\(x + 8\)[/tex] must be positive.
For [tex]\(x = -\frac{3}{5}\)[/tex], both terms are positive:
[tex]\[ 5\left(-\frac{3}{5}\right)^2 + 2 > 0 \quad \text{and} \quad -\frac{3}{5} + 8 > 0 \][/tex]
For [tex]\(x = 2\)[/tex], both terms are positive:
[tex]\[ 5(2)^2 + 2 > 0 \quad \text{and} \quad 2 + 8 > 0 \][/tex]
Thus, both solutions are valid. So the completed answer is:
[tex]\[ \boxed{-\frac{3}{5} \text{ and } 2} \][/tex]
### Equation 3: [tex]\(e^{x^2} = e^{4x + 5}\)[/tex]
Since the bases are the same, we equate the exponents:
[tex]\[ x^2 = 4x + 5 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - 4x - 5 = 0 \][/tex]
Factor:
[tex]\[ (x - 5)(x + 1) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = 5 \text{ and } x = -1 \][/tex]
So the completed answer is:
[tex]\[ \boxed{-1 \text{ and } 5} \][/tex]
### Equation 4: [tex]\(\ln (x + 5) = \ln (x - 1) + \ln (x + 1)\)[/tex]
Use the properties of logarithms to combine the right-hand side:
[tex]\[ \ln (x + 5) = \ln ((x - 1)(x + 1)) \][/tex]
Thus:
[tex]\[ x + 5 = (x - 1)(x + 1) \][/tex]
Expand and simplify:
[tex]\[ x + 5 = x^2 - 1 \][/tex]
Rearrange to form a quadratic equation:
[tex]\[ x^2 - x - 6 = 0 \][/tex]
Factor:
[tex]\[ (x - \frac{6}{5})(x + \frac{1}{5}) = 0 \][/tex]
Thus, the solutions are:
[tex]\[ x = \frac{6}{5} \text{ and } x = -\frac{1}{5} \][/tex]
Check for the domain: With natural logarithms, the arguments must be positive.
For [tex]\(x = \frac{6}{5}\)[/tex], [tex]\(x + 5 > 0, x - 1 > 0, x + 1 > 0\)[/tex].
For [tex]\(x = -\frac{1}{5}\)[/tex], [tex]\(x + 5 > 0\)[/tex], but [tex]\(x - 1 < 0\)[/tex], making it invalid.
Thus, the valid solution is:
[tex]\[ \boxed{\frac{6}{5}} \][/tex]
Matching each solution to its respective equation:
[tex]\[ \begin{array}{c} \log (x - 1) + \log 5x = 2 \quad \rightarrow \quad \boxed{\text{only 5}} \\ \log_4(5x^2 + 2) = \log_4(x + 8) \quad \rightarrow \quad \boxed{-\frac{3}{5} \text{ and } 2} \\ e^{x^2} = e^{4x + 5} \quad \rightarrow \quad \boxed{-1 \text{ and } 5} \\ \ln (x + 5) = \ln (x - 1) + \ln (x + 1) \quad \rightarrow \quad \boxed{\frac{6}{5}} \end{array} \][/tex]
